To find:
Formula Used:
Let y = (1 – 9x) … (1)
Differentiating with respect to x,
i.e., dy = -9 dx
Substituting in the equation to evaluate,
Simplifying and substituting the value of y from (1),
Therefore,
Evaluate:
To find:
Formula Used:
Given equation =
… (1)
Here
Therefore, (1) becomes
Therefore,
Evaluate:
To find:
Formula Used:
Rewriting the given equation,
Here a = 4
Therefore,
Evaluate:
To find:
Formula Used:
Rewriting the given equation,
Here
Therefore,
Evaluate:
To find:
Formula Used:
Rewriting the given equation,
Here a = 5
Therefore,
Evaluate:
To find:
Formula Used:
Rewriting the given equation,
Here
Therefore,
Evaluate:
To find:
Formula Used:
Given equation can be rewritten as the following:
Here a = 2,
Therefore,
Evaluate:
To find:
Formula Used:
Given equation can be rewritten as the following:
Here ,
Therefore,
Evaluate:
To find:
Formula Used:
Let y = ex … (1)
Differentiating both sides, we get
dy = ex dx
Substituting in given equation,
⇒ tan-1 y
From (1),
⇒ tan-1 (ex)
Therefore,
Evaluate:
To find:
Formula Used:
Let y = cos x … (1)
Differentiating both sides, we get
dy = –sin x dx
Substituting in given equation,
⇒ – tan-1 y
From (1),
⇒ –tan-1 (cos x)
Therefore,
Evaluate:
To find:
Formula Used:
Let y = sin x … (1)
Differentiating both sides, we get
dy = cos x dx
Substituting in given equation,
⇒ tan-1 y
From (1),
⇒ tan-1 (sin x)
Therefore,
Evaluate:
To find:
Formula Used:
Let y = x6 … (1)
Differentiating both sides, we get
dy = 6x5 dx
Substituting in given equation,
From (1),
Therefore,
Evaluate:
To find:
Formula Used:
Let y = x4 … (1)
Differentiating both sides, we get
dy = 4x3 dx
Substituting in given equation,
From (1),
Therefore,
Evaluate:
To find:
Formula Used:
Given equation is:
… (1)
Let y = ex … (1)
Differentiating both sides, we get
dy = ex dx
Substituting in (1),
⇒ tan-1 y
From (1),
⇒ tan-1 (ex)
Therefore,
Evaluate:
To find:
Formula Used:
Let y = x2 … (1)
Differentiating both sides, we get
dy = 2x dx
Substituting in given equation,
Here a = 1,
From (1),
Therefore,
Evaluate:
To find:
Formula Used:
Let y = x3 … (1)
Differentiating both sides, we get
dy = 3x2 dx
Substituting in given equation,
From (1),
Therefore,
Evaluate:
To find:
Formula Used:
Rewriting the given equation,
… (1)
Let y = x + 2 … (2)
Differentiating both sides,
dy = dx
Substituting in (1),
Here a = 2,
From (2),
Therefore,
Evaluate:
To find:
Formula Used:
Rewriting the given equation,
… (1)
Let y = 2x – 1 … (2)
Differentiating both sides,
dy = 2dx
Substituting in (1),
Here a = √2,
From (2),
Therefore,
Evaluate:
To find:
Formula Used:
Rewriting the given equation,
… (1)
Let … (2)
Differentiating both sides,
dy = √2 dx
Substituting in (1),
Here
From (2),
Therefore,
Evaluate:
To find:
Formula Used:
Rewriting the given equation,
… (1)
Let … (2)
Differentiating both sides,
dy = √2 dx
Substituting in (1),
Here
From (2),
Therefore,
Evaluate:
To find:
Formula Used:
Rewriting the given equation,
… (1)
Let y = x + 1 … (2)
Differentiating both sides wrt x,
dy = dx
Substituting in (1),
Here a = 2,
From (2),
Therefore,
Evaluate:
To find:
Formula Used:
1.
2.
Using partial fractions,
x = A (2x + 3) + B
Equating the coefficients of x,
1 = 2A
Also, 0 = 3A + B
Therefore, the given equation becomes,
Therefore,
Evaluate:
To find:
Formula Used:
1.
2.
Using partial fractions,
x – 3 = A (2x + 2) + B
Equating the coefficients of x,
1 = 2A
Also, -3 = 2A + B
⇒ B = -4
Substituting in the given equation,
Therefore,
Evaluate:
To find:
Formula Used:
1.
2.
Using partial fractions,
2x – 3 = A(2x + 3) + B
Equating the coefficients of x,
2 = 2A
A = 1
Also, -3 = 3A + B
⇒ B = -6
Substituting in the given equation,
… (1)
Let I
Here a
… (2)
Substituting (2) in (1),
Therefore,
Evaluate:
To find:
Formula Used:
1.
2.
Given equation can be rewritten as following:
Let I … (2)
Using partial fractions,
6x – 3 = A (2x + 6) + B
Equating the coefficients of x,
6 = 2A
A = 3
Also, -3 = 6A + B
⇒ B = -21
Substituting in (1),
I
Therefore,
Evaluate:
To find:
Formula Used:
1.
2.
Using partial fractions,
2x – 1 = A (4x + 2) + B
Equating the coefficients of x,
2 = 4A
A
Also, -1 = 2A + B
⇒ B = -2
Substituting in the given equation,
Let I … (1)
Here
⇒ 2tan-1(2x + 1) + C
Substituting in (1) and combining with original equation,
Therefore,
Evaluate:
To find:
Formula Used:
1.
2.
Rewriting the given equation,
Using partial fractions,
3x – 1 = A (6x + 4) + B
Equating the coefficients of x,
3 = 6A
A
Also, -1 = 4A + B
⇒ B = -3
Substituting in the original equation,
Let I
Here
Substituting in (1) and combining with original equation,
Therefore,
Evaluate:
To find:
Formula Used:
1.
2.
Rewriting the given equation,
Using partial fractions,
x = A (2x – 1) + B
Equating the coefficients of x,
1 = 2A
A
Also, 0 = -A + B
B
Substituting in the original equation,
Let I
Here
Substituting for I and combining with the original equation,
Therefore,
or
Evaluate:
To find:
Formula Used:
1.
2. sec2 x = 1 + tan2 x
Dividing the given equation by cos2x in the numerator and denominator gives us,
… (1)
Let y = tan x
dy = sec2 x dx … (2)
Also, y2 = tan2 x
i.e., y2 = sec2 x – 1
sec2 x = y2 + 1 … (3)
Substituting (2) and (3) in (1),
Since y = tan x,
Therefore,
Evaluate:
To find:
Formula Used:
1.
2. sec2 x = 1 + tan2 x
Dividing the given equation by cos2x in the numerator and denominator gives us,
… (1)
Let y = tan x
dy = sec2 x dx … (2)
Also, y2 = tan2 x
i.e., y2 = sec2 x – 1
sec2 x = y2 + 1 … (3)
Substituting (2) and (3) in (1),
Since y = tan x,
Therefore,
Evaluate:
To find:
Formula Used:
1. Sec2 x = 1 + tan2 x
2.
Dividing by cos2 x in the numerator and denominator,
Let y = tan x
dy = sec2 x dx
Therefore,
Since y = tan x,
Therefore,
Evaluate:
To find:
Formula Used:
1. sec2 x = 1 + tan2 x
2.
Dividing by cos2 x in the numerator and denominator,
Let y = tan x
dy = sec2 x dx
Therefore,
Since y = tan x,
Therefore,
Evaluate:
To find:
Formula Used:
1. sec2 x = 1 + tan2 x
2.
Dividing by cos2 x in the numerator and denominator,
Let y = tan x
dy = sec2 x dx
Therefore,
Since y = tan x,
Therefore,
Evaluate:
To find:
Formula Used:
1. sec2 x = 1 + tan2 x
2.
Dividing by cos2 x in the numerator and denominator,
Let y = tan x
dy = sec2 x dx
Therefore,
⇒ log |y + 2| + C
Since y = tan x,
⇒ log |tan x + 2| + C
Therefore,
Evaluate:
To find:
Formula Used:
1. sec2 x = 1 + tan2 x
2.
3. sin 2x = 2 sin x cos x
Rewriting the given equation,
Dividing by cos4 x in the numerator and denominator,
Let y = tan x
dy = sec2 x dx
Therefore,
Let z = y2
dz = 2y dy
⇒ tan-1 z + C
Since z = y2,
⇒ tan-1(y2) + C
Since y = tan x,
⇒ tan-1(tan2 x) + C
Therefore,
Evaluate:
To find:
Formula Used:
1. sec2 x = 1 + tan2 x
2.
3. sin 2x = 2 sin x cos x
Rewriting the given equation,
Let y = sin ϕ
dy = cos ϕ dϕ
Substituting in the original equation,
… (1)
Using partial fraction,
4y – 1 = A (2y - 4) + B
Equating the coefficients of y,
4 = 2A
A = 2
Also, -1 = -4A + B
B = 7
Substituting in (1),
⇒ 2 log |y2 – 4y + 5| + 7 tan-1(y – 2) + C
But y = sin ϕ
⇒ 2 log |sin2ϕ – 4 sin ϕ + 5| + 7 tan-1(sin ϕ – 2) + C
Therefore,
Evaluate:
To find:
Formula Used:
1. sec2 x = 1 + tan2 x
2.
Dividing by cos2 x in the numerator and denominator,
Let y = tan x
dy = sec2 x dx
Therefore,
… (1)
Let
1 = A (2y + 1) + B(y – 2)
When y = 0,
1 = A – 2B … (2)
When y = 1,
1 = 3A – B ⇒ 2 = 6A – 2B … (3)
Solving (2) and (3),
1 = 5A
A
So, B
(1) becomes,
Since y = tan x,
Therefore,
Evaluate:
To find:
Formula used:
On dividing by x2 in the numerator and denominator of the given equation,
Let
Differentiating wrt x,
Substituting in the original equation,
Substituting for and taking reciprocal of the value within logarithm, we get
Therefore,
Evaluate:
To find:
Formula used:
On dividing by x2 in the numerator and denominator of the given equation,
Let
Differentiating wrt x,
Substituting in the original equation,
Substituting for
Therefore,
Evaluate:
To find:
Formula used:
1. sec2 x = 1 + tan2x
2.
Dividing by cos4 x in the numerator and denominator of the given equation,
Let y = tan x
dy = sec2 x dx
Substituting in the original equation,
Dividing by y2 in the numerator and denominator,
Let
Therefore,
Substituting for z,
Substituting for y = tan x,
Therefore,
Evaluate:
Formula to be used - where c is the integrating constant
, c being the integrating constant
Formula to be used - where c is the integrating constant
, c being the integrating constant
Evaluate:
Formula to be used - where c is the integrating constant
, c being the integrating constant
Evaluate:
Formula to be used - where c is the integrating constant
, c being the integrating constant
Evaluate:
Formula to be used - where c is the integrating constant
, c being the integrating constant
Evaluate:
Formula to be used - where c is the integrating constant
, c being the integrating constant
Evaluate:
Formula to be used - where c is the integrating constant
, c being the integrating constant
Evaluate:
Formula to be used - where c is the integrating constant
, c being the integrating constant
Evaluate:
Formula to be used - where c is the integrating constant
, c being the integrating constant
Evaluate:
Tip – d(x2) = 2xdx i.e. xdx = (1/2)×d(x2)
Formula to be used - where c is the integrating constant
, c being the integrating constant
Evaluate:
Tip – d(x3) = 3x2dx so, d(4x3) = 4×3x2dx i.e 3x2dx = (1/4)d(2x3)
Formula to be used - where c is the integrating constant
, c being the integrating constant
Evaluate:
Tip – d(tanx) = sec2xdx
Formula to be used - where c is the integrating constant
, c being the integrating constant
Evaluate:
Tip – d(cosx) = - sinxdx i.e. sinxdx = - d(cosx)
Formula to be used - where c is the integrating constant
, c being the integrating constant
Evaluate:
Tip – d(sinx) = cosxdx so, d(3sinx) = 3cosxdx i.e. cosxdx = (1/3)d(3sinx)
Formula to be used - where c is the integrating constant
, c being the integrating constant
Evaluate:
Tip – d(ex) = exdx
Formula to be used - where c is the integrating constant
, c being the integrating constant
Evaluate:
Tip – d(ex) = exdx
Formula to be used - where c is the integrating constant
, c being the integrating constant
Evaluate:
Formula to be used - where c is the integrating constant
Tip – Assuming e - (x/2) = a, - (1/2) e - (x/2)dx = da i.e. e - (x/2)dx = - 2da
, c being the integrating constant
Evaluate:
Tip – Taking ,
and
i.e
Formula to be used -
, c being the integrating constant
Evaluate:
Formula to be used - where c is the integrating constant
, c being the integrating constant
Evaluate:
Tip – d(2 - x) = - dx i.e. dx = - d(2 - x)
Formula to be used - where c is the integrating constant
, c being the integrating constant
Evaluate:
Formula to be used - where c is the integrating constant
, c being the integrating constant
Evaluate:
Formula to be used - where c is the integrating constant
, c being the integrating constant
Evaluate:
Formula to be used - where c is the integrating constant
, c being the integrating constant
Evaluate:
Formula to be used - where c is the integrating constant
, c being the integrating constant
Evaluate:
Formula to be used - where c is the integrating constant
, c being the integrating constant
Evaluate:
Formula to be used - where c is the integrating constant
, c being the integrating constant
Evaluate:
Formula to be used - where c is the integrating constant
, c being the integrating constant
Evaluate:
Formula to be used - where c is the integrating constant
, c being the integrating constant
Evaluate:
Formula to be used - where c is the integrating constant
, c being the integrating constant
Evaluate:
Formula to be used - where c is the integrating constant
, c being the integrating constant
Evaluate:
Formula to be used - where c is the integrating constant
, c being the integrating constant
Evaluate:
Formula to be used - where c is the integrating constant
, c being the integrating constant
Evaluate:
Formula to be used - where c is the integrating constant
, c being the integrating constant
Evaluate:
Formula to be used - where c is the integrating constant
, c being the integrating constant
Evaluate:
Tip – d(x3) = 3x2dx i.e. x2dx = (1/3)d(x3)
Formula to be used - where c is the integrating constant
, c being the integrating constant
Evaluate:
Formula to be used - where c is the integrating constant
Tip – Assuming x2 + x + 1 = a2, (2x + 1)dx = 2ada
, c is the integrating constant
Evaluate:
Formula to be used - where c is the integrating constant
Tip – Assuming x2 + 4x + 10 = a2, (2x + 4)dx = 2ada
, c is the integrating constant
Evaluate:
Formula to be used - where c is the integrating constant
Tip – Assuming 2x2 + 2x - 3 = a2, (4x + 2)dx = 2ada
, c is the integrating constant
Evaluate:
Formula to be used - where c is the integrating constant
Tip – Assuming 2 + x - x2 = a2, (1 - 2x)dx = 2ada
, c is the integrating constant
Evaluate:
Formula to be used - where c is the integrating constant
Tip – Assuming 2x2 + 2x - 3 = a2, (4x + 2)dx = 2ada
, c is the integrating constant
Evaluate:
Formula to be used - where c is the integrating constant
Tip – Assuming 5 - 2x - x2 = a2, ( - 2 - 2x)dx = 2ada i.e. (x + 1)dx = - ada
, c is the integrating constant
Evaluate:
Formula to be used - where c is the integrating constant
Tip – Assuming 6 + x - 2x2 = a2, (1 - 4x)dx = 2ada i.e. (4x - 1)dx = - 2ada
, c is the integrating constant
Evaluate:
Formula to be used - where c is the integrating constant
Tip – Taking x2 + x = a2 , (2x + 1)dx = 2ada
, c is the integrating constant
Evaluate:
Formula to be used - where c is the integrating constant
Tip – Taking x2 + 5x + 6 = a2 , (2x + 5)dx = 2ada
, c is the integrating constant
Evaluate the following integrals:
To Find :
Now, can be written as
Formula Used: = + + C
Since is of the form ,
Hence, = + + C
= + + C
= + 2 + C
Therefore, = + 2 + C
To Find :
Now, can be written as
Formula Used: = + + C
Since is of the form ,
Hence, = + + C
= + + C
= + + C
Therefore, = + + C
Evaluate the following integrals:
To Find :
Now, can be written as
Formula Used: = log |x +|+ C
Since is of the form ,
Hence, = log |x +|+ C
= log |x +|+ C
= log |x +|+ C
Therefore, = log |x +|+ C
Evaluate the following integrals:
To Find :
Now, can be written as
Formula Used: = log |x +|+ C
Since is of the form ,
Hence, = log |x +|+C
= log |x +|+ C
= log |x +|+ C
Therefore, = log |x +|+ C
Evaluate the following integrals:
To Find :
Now, can be written as
Formula Used: = log |x +|+ C
Since is of the form ,
Hence, = log |x +|+ C
= log |x +|+ C
Therefore, = log |x +|+ C
Evaluate the following integrals:
To Find :
Now, can be written as
Formula Used: = log |x +|+ C
Since is of the form ,
Hence, = log |2x +|+ C
= log |2x + | + C
= log |2x +|+ C
Therefore, = log |2x +|+ C
Evaluate the following integrals:
To Find :
Now, can be written as
Formula Used: = log |x +|+ C
Since is of the form ,
Hence, = log |x +|+ C
= log |x + | + C
= log |x +|+ C
Therefore, = log |x +|+ C
Evaluate the following integrals:
To Find :
Now, let sin x = t
cosx dx = dt
Therefore, can be written as
Formula Used: = + + C
Since , is in the form of with t as a variable instead of x .
= + + C
= + + C
Now since sin x = t and cosx dx = dt
= + + C
Evaluate the following integrals:
To Find :
Now, can be written as
i.e.,
Here , let x – 2 = y dx = dy
Therefore, can be written as
Formula Used: = log |x +|+ C
Since is of the form with change in variable.
= log |y +|+ C
= log |y +|+ C
= 2 log |y +|+ C
Since , x – 2 = y and dx = dy
= 2 log |(x-2) +|+C Therefore,
= 2 log |(x - 2) +|+ C
Evaluate the following integrals:
To Find :
Now, can be written as
i.e,
Here , let x + 3 = y dx = dy
Therefore, can be written as
Formula Used: = log |x +|+ C
Since is of the form with change in variable.
= log |y +|+ C
= log |y +|+ C
Since , x + 3 = y and dx = dy
= log |(x + 3) +|+C
Therefore,
= log |(x + 3) +|+C
Evaluate the following integrals:
To Find : dx
Now, can be written as
i.e,
Let x – 1 = y dx = dy
Therefore , becomes
Formula Used: = + + C
Since is of the form with change in variable,
Hence = + + C
= + + C
Here we have x – 1 = y and dx = dy
= + + C
Therefore , = + + C
Evaluate the following integrals:
To Find : dx
Now, can be written as
i.e,
Let x + 2 = y dx = dy
Therefore , becomes
Formula Used: = + + C
Since is of the form with change in variable,
Hence = + + C
= + + C
Here we have x + 2 = y and dx = dy
= + + C
Therefore , = + + C
Evaluate the following integrals:
To Find : dx
Now, can be written as
i.e,
Let x - a = y dx = dy
Therefore , becomes
Formula Used: = + + C
Since is of the form with change in variable,
Hence = + + C
= + + C
Here we have x - a = y and dx = dy
= + + C
Therefore , = + + C
Evaluate the following integrals:
To Find :
Now , consider =
=
=
=
Let x + = y dx = dy
Hence becomes
Formula Used: = log |x +|+ C
Now consider which is in the form of with change in variable.
= log |y +|+ C
= log |y +|+ C
Since x + = y and dx = dy
= log | +|+ C
Now , = log | +|+ C
Therefore,
= log | +|+ C
Evaluate the following integrals:
To Find :
Now, can be written as
i.e,
Here , let x + = y dx = dy
Therefore, can be written as
Formula Used: = log |x +|+ C
Since is of the form with change in variable.
= log |y +|+ C
= log |y +|+ C
Since , x + = y and dx = dy
= log |(x + ) +|+C
Therefore,
= log |x + +|+C
Evaluate the following integrals:
To Find :
Now, can be written as
i.e,
Here , let x + = y dx = dy
Therefore, can be written as
Formula Used: = log |x +|+ C
Since is of the form with change in variable.
= log |y +|+ C
= log |y +|+ C
Since , x + = y and dx = dy
= log |(x + ) +|+C
Therefore,
= log |x + +|+C
Evaluate the following integrals:
To Find :
Now, let be written as (2x – 4) -1 and split
Therefore ,
=
=
Now solving,
Let = u dx =
Thus, becomes
Now , = = =
=
Now solving,
=
=
Let x – 2 = y dx = dy
Then becomes
Formula Used: = log |x +|+ C
Since is in the form of with change in variable.
Hence = log |y +|+ C
= log |y +|+ C
Now, since x – 2 = y and dx = dy
= log |(x-2) +|+ C
Hence = log |(x-2) +|+ C
Therefore , =
log |(x-2) +|+ C
=
log |x - 2 +|+ C
Evaluate the following integrals:
To Find :
Now, let x + 2 be written as (2x + 1) + and split
Therefore ,
=
=
Now solving,
Let = u dx =
Thus, becomes
Now , = = ) =
=
Now solving ,
Now, can be written as
i.e,
Here , let x + = y dx = dy
Therefore, can be written as
Formula Used: = log |x +|+ C
Since is of the form with change in variable.
= log |y +|+ C
= log |y +|+ C
Since , x + = y and dx = dy
= log |(x + ) +|+C
Therefore,
= log |x + +|+C
Hence ,
= + log |(x + )+|+C
Therefore , = + log |(x + )+|+C
Evaluate the following integrals:
To Find :
Now, let x - 5 be written as (2x + 1) - and split
Therefore ,
=
=
Now solving,
Let = u dx =
Thus, becomes
Now , = = ) =
=
Now solving,
Now, can be written as
i.e,
Here , let x + = y dx = dy
Therefore, can be written as
Formula Used: = log |x +|+ C
Since is of the form with change in variable.
= log |y +|+ C
= log |y +|+ C
Since , x + = y and dx = dy
= log |(x + ) +|+C
Therefore,
= log |x + +|+C
Now ,
= - log |x + +|+C
Therefore ,
= - log |x + +|+C
Evaluate the following integrals:
To Find :
Now, let 4x + 1 be written as 2(2x - 1) + 3 and split
Therefore ,
=
=
Now solving,
Let = u dx =
Thus, becomes
Now , = = ) =
=
Now solving,
Now, can be written as
i.e,
Here , let x - = y dx = dy
Therefore, can be written as
Formula Used: = log |x +|+ C
Since is of the form with change in variable.
= log |y +|+ C
= log |y +|+ C
Since , x - = y and dx = dy
= log |(x - ) +|+C
Therefore,
= log |x - +|+C
Hence ,
= + log |x - +|+C
Therefore ,
= + log |x - +|+C
Evaluate the following integrals:
To Find :
Now, can be written as
=
=
Now solving,
Let = u dx =
Thus, becomes
Now , = = ) =
=
Now solving,
Now, can be written as
Formula Used: = log |x +|+ C
Since is of the form .
= log |x +|+ C
= log |x +|+ C
Therefore,
= + log |x +|+ C
Hence ,
= + log |x +|+ C
Evaluate the following integrals:
To Find :
Now, let x be written as - (1 - 2x) and split
Therefore ,
=
= +
Now solving,
Let = u dx =
Thus,becomes
Now , = = ) =
=
Now solving,
can be written as
i.e, =
let 2x – 1 = y dx =
Therefore , becomes
Formula Used: = + + C
Since is of the form with change in variable .
Hence, = + + C
= + + C
Since , 2x – 1 = y and dx =
Therefore,
= + + C
i.e, = + + C
hence , = + = = + + + C
Evaluate the following integrals:
To Find :
Now, let 2x - 5 be written as (2x – 3) -2 and split
Therefore ,
=
= -
Now solving,
Let = u dx =
Thus,becomes
Now , = = ) =
=
Now solving,
can be written as
i.e,
let x – = y dx = dy
Therefore , becomes
Formula Used: = + + C
Since is of the form with change in variable .
Hence, = + + C
= + + C
Since , x – = y and dx = dy
Therefore,
= + + C
i.e, = + + C
hence ,
= - = - - + C
Evaluate the following integrals:
To Find :
Now, let 6x + 5 be written as - (1 - 4x)and split
Therefore ,
=
= +
Now solving,
Let = u dx =
Thus,becomes
Now , = = ) =
=
Now solving,
can be written as
i.e,
let = y dx =
Therefore , becomes
Formula Used: = + + C
Since is of the form with change in variable .
Hence, = + + C
= + + C
Since , = y and dx =
Therefore,
= + + C
i.e, = + + C
hence ,
= + = + + + C
Evaluate the following integrals:
To Find :
Now, let x + 1 be written as - (-2x - 1) and split
Therefore ,
=
= +
Now solving,
Let = u dx =
Thus,becomes
Now , = = ) =
=
Now solving,
can be written as
i.e,
let = y dx = dy
Therefore , becomes
Formula Used: = + + C
Since is of the form with change in variable .
Hence, = + + C
= + + C
Since , = y and dx = dy
Therefore,
= + + C
i.e, = + + C
hence ,
= + = + + + C
Evaluate the following integrals:
To Find :
Now, let x - 3 be written as (2x + 3) - and split
Therefore ,
=
= -
Now solving,
Let = u dx =
Thus,becomes
Now , = = ) =
=
Now solving,
can be written as
i.e,
let = y dx = dy
Therefore, can be written as
Formula Used: = log |x +|+ C
Since is of the form with change in variable.
= log |y +|+ C
= log |y +|+ C
Since , x + = y and dx = dy
= log |(x + ) +|+C
Therefore,
= log |x + +|+C
Hence ,
= - = log |x + +|+C