Integration Using Partial Fractions

Let ,

Putting

Which implies A(x+2) + Bx = 1, putting x+2=0

Therefore x=-2,

And B = -0.5

Now put x=0, A= � ,

From equation (1), we get

Let ,

Putting

Which implies 2x=1 = A(x-3) + B(x+2)

Now put x-3=0, x=3

2×3+1=A(0 )+B 3+2)

So

Now put x+2=0 ,x=-2

-4+1=A(-2-3) + B(0)

So

From equation (1), we get ,

Let

Putting

Which implies A(3-2x)+B(x+2)=x

Now put 3-2x=0

Therefore,

Now put x+2=0

Therefore, x=-2

A(7)+B(0)=-2

Now From equation (1) we get

Let

Putting

Which implies,

A(x-2)(x-4)+Bx(x-4)+Cx(x-2)=1

Now put x-2=0

Therefore, x=2

A(0)+B×2(2-4)+C(0)=1

B×2(-2)=1

Now put x-4=0

Therefore, x=4

A(0)+B×(0)+C×4(4-2)=1

C×4(2)=1

Now put x=0

A(0-2)(0-4)+B(0)+C(0)=1

Now From equation (1) we get

Let

Putting

Which implies,

A(x+2)(x-2)+B(x-1)(x-3)+C(x-1)(x+2)=2x-1

Now put x+2=0

Therefore, x=-2

A(0)+B(-2-1)(-2-3)+C(0)=2x-2-1

B(-3)(-5)=-5

Now put x-3=0

Therefore, x=3

A(0)+B(0)+C(2)(5)=5

Now put x-1=0

Therefore, x=1

A(3)(-2)+B(0)+C(0)=1

Now From equation (1) we get,

Let

Putting

Which implies,

A(x+1)(2x+3)+B(x-1)(2x+3)+C(x-1)(x+1)=2x-3

Now put x+1=0

Therefore, x=-1

A(0)+B(-1-1)(-2+3)+C(0)=-2-3

Now put x-1=0

Therefore, x=1

A(2)(2+3)+B(0)+C(0)=-1

Now put 2x+3=0

Therefore,

.Now From equation (1) we get,

Let

Putting

Which implies,

A(x+1)+B(x-2)=2x+5

Now put x+1=0

Therefore, x=-1

A(0)+B(-1-2)=3

B=-1

Now put x-2=0

Therefore, x=2

A(2+1)+B(0)=2×2+5=9

A=3

Now From equation (1) we get,

Let

Which implies

Therefore, I=x+I₁

Where, I₁=

Putting

Which implies,

A(x+2)+B(x+1)=2x+1

Now put x+2=0

Therefore, x=-2

A(0)+B(-1)=-4+1

B=3

Now put x+1=0

Therefore, x=-1

A(-1+2)+B(0)=-2+1

A=-1

Now From equation (1) we get,

Let

Let

Let

So

Therefore

Putting x²-4=t

2xdx = dt

Putting the value of I₁ in I,

Let

Put

A(x-1)+B(x+2)=5x+1

Now put x-1=0

Therefore, x=1

A(0)+B(1+2)=5+1=6

B=2

Now put x+2=0

Therefore, x=-2

A(-2-1)+B(0)=5×(-2)+1

A=3

Now From equation (1) we get,

Therefore,

Let

Where,

Putting ……….(2)

A(x-2)+B(x-1)=7x-6

Now put x-2=0

Therefore, x=2

A(0)+B(2-1)=7×2-6

B=8

Now put x-1=0

Therefore, x=1

A(1-2)+B(0)=7-6=1

A=-1

Now From equation (2) we get,

Now From equation (1) we get,

Let

Let

Putting

A(4+x)+B(1-x)=2x+1

Now put 1-x=0

Therefore, x=1

A(5)+B(0)=3

Now put 4+x=0

Therefore, x=-4

A(0)+B(5)=-8+1=-7

Now From equation (1) we get,

Put x²=t

2xdx=dt

Let

Putting t=sin x

dt=cos x dx

Now putting,

A(2+t)+B(1+t)=1

Now put t+1=0

Therefore, t=-1

A(2-1)+B(0)=1

A=1

Now put t+2=0

Therefore, t=-2

A(0)+B(-2+1)=1

B=-1

Now From equation (1) we get,

So,

Let

Putting t=tanx

dt=sec²xdx

Now putting,

A(3+t)+B(2+t)=1

Now put t+2=0

Therefore, t=-2

A(3-2)+B(0)=1

A=1

Now put t+3=0

Therefore, t=-3

A(0)+B(2-3)=1

B=-1

Now From equation (1) we get,

So,

Let

Putting t=cos x

dt=-sin x dx

Now putting,

A(t-2)+B(t+1)=-t

Now put t-2=0

Therefore, t=2

A(0)+B(2+1)=-2

Now put t+1=0

Therefore, t=-1

A(-1-2)+B(0)=1

Now From equation (1) we get,

So,

Let

Putting t=e^x

dt=e^x dx

Now putting,

A(3+t)+B(2+t)=1

Now put t+2=0

Therefore, t=-2

A(3-2)+B(0)=1

A=1

Now put t+3=0

Therefore, t=-3

A(0)+B(2-3)=1

B=-1

Now From equation (1) we get,

Let

Putting t=e^x

dt=e^x dx

Now putting,

A(t+1)(t-3)+B(t-1)(t-3)+C(t-1)(t+1)=1

Now put t+1=0

Therefore, t=-1

A(0)+B(-1-1)(-1-3)+C(0)=1

B(-2)(-4)=1

Now put t-1=0

Therefore, t=1

A(1+1)(1-3)+B(0)+C(0)=1

Now put t-3=0

Therefore, t=3

A(0)+B(0)+C(3-1)(3+1)=1

Now From equation (1) we get,

Let

Putting t=log x

dt=dx/x

Now putting,

A(t+1)+B(2t-3)=2t

Now put 2t-3=0

Therefore,

Now put t+1=0

Therefore, t=-1

A(0)+B(-2-3)=-2

Now From equation (1) we get,

Let

Putting t=cot x

dt=-cosec²xdx

Let

Putting t=tan x

dt=sec²xdx

Now putting,

A(t²+4)+(Bt + C)t=1

Putting t=0,

A(0+4)× B(0)=1

By equating the coefficients of t² and constant here,

A+B=0

Now From equation (1) we get,

Let

Putting t=sin x

dt=cos x dx

Now putting,

A(2+t)+B(1+t)=2t

Now put t+2=0

Therefore, t=-2

A(0)+B(1-2)=-4

B=4

Now put t+1=0

Therefore, t=-1

A(2-)+B(0)=-2

A=-2

Now from equation (1), we get,

So,

Let

Putting t=e^x

dt=e^xdx

Now putting,

A(t-1)+Bt=1

Now put t-1=0

Therefore, t=1

A(0)+B(1) =1

B=1

Now put t=0

A(0-1)+B(0)=1

A=-1

Now From equation (1) we get,

Let

Putting t=x⁴

dt=4x³dx

Now putting,

A(t-1)+Bt=1

Now put t-1=0

Therefore, t=1

A(0)+B(1) =1

B=1

Now put t=0

A(0-1)+B(0)=1

A=-1

Now From equation (1) we get,

Let

Where ……..(2)

Now putting,

A(2x-1)+Bx=1

Putting 2x-1=0

B=2

Putting x=0,

A(0-1)+B(0)=1

A=-1

From equation (2), we get,

From equation (1),

Let

Now putting,

A(x+1)²+B(x+2)(x+1)+C(x+2)=x²+x+1

Now put x+1=0

Therefore, x=-1

A(0)+B(0)+C(-1+2) =1-1+1=1

C=1

Now put x+2=0

Therefore, x=-2

A(-2+1)²+B(0)+C(0) =4-2+1=3

A=3

Equating the coefficient of x²,A+B=1

3+B=1

B=-2

Form equation (1),we get,

So,

Let

Now putting,

A(x-3)²+B(x+2)(x-3)+C(x+2)=2x+9

Now put x-3=0

Therefore, x=3

A(0)+B(0)+C(3+2) =6+9=15

C=3

Now put x+2=0

Therefore, x=-2

A(-2-3)²+B(0)+C(0) = -4+9=5

Equating the coefficient of x²,we get,

A+B=0

From equation (1), we get,

Let

Now putting,

A(x-1)²+B(x+3)(x-1)+C(x+3)=x²+1

Now put x-1=0

Therefore, x=1

A(0)+B(0)+C(4) =2

Now put x+3=0

Therefore, x=-3

A(-3-1)²+B(0)+C(0) =9+1=10

By equating the coefficient of x², we get, A+B=1

From equation (1), we get,

Let

Now putting,

A(x-1)²+B(x-3)(x-1)+C(x-3)=x²+1

Putting x-1=0,

X=1

A(0)+B(0)+C(1-3)=1+1

C=-1

Putting x-3=0,

X=3

A(3-1)²+B(0)+C(0)=9+1

A(4)=10

Equating the coefficient of x²

A+B=1

From (i)

Let

Now putting,

A(x²+1)+(Bx+C)(x+2) = x²+x+1

Ax²+A+Bx²+Cx+2Bx+2C = x²+x+1

(A+B)x²+(C+2B)x+(A+2C) = x²+x+1

Equating coefficients A+B=1…….(i)

A+2C=1

A=1-2C……(ii)

2B+C=1

2B=1-C

2-4C+1-C=2

3-5C=2

-5C=-1

And

Let

Now putting,

A(2x+1)+B = 2x

Putting 2x+1=0,

A(0)+B=-1

B=-1

By equating the coefficient of x,

2A=2

A=1

From equation (1),we get,

Let

Now putting,

A(x-2)²+B(x+2)(x-2)+C(x+2)=3x+1

Putting x-2=0,

X=2

A(0)+B(0)+C(2+1)=3×2+1

Putting x+2=0,

X=-2

A(-4)²+B(0)+C(0)=-6+1=-5

By equation the coefficient of x², we get, A+B=0

Let

Now putting,

Ax²+(Bx + C)(3x+8) = 5x+8

Putting 3x+8=0,

By equating the coefficient of x² and constant term,

A+3B=0

8C=8

C=1

From equation (1), we get,

Putting x+2=0,

X=-2

A(-4)²+B(0)+C(0)=-6+1=-5

Let

Now putting,

A(x-1)²+B(2x-3)(x-1)+C(2x-3) = 5x²-18x+17

Putting x-1=0,

X=1

A(0)+B(0)+C(2-3)=5-18+17

C(-1)=4

Putting 2x-3=0,

A=5

By equating the coefficient of x², we get ,

A+2B=5

5+2B=5

2B=0

B=0

From equation (1), we get,

Let

Now putting,

A(x²+4)+(Bx+C)(x+2)= 8

Putting x+2=0,

X=-2

A(4+4)+0=8

A=1

By equating the coefficient of x² and constant term, A+B=0

1+B=0

B=-1

4A+2C=8

4×1+2C=8

2C=4

C=2

From equation (1),we get,

Let

Now putting,

A(x²+1)+(Bx+C)(x-1)=3x+5

Putting x-1=0,

X=1

A(2)+B(0)=3+5=8

A=4

By equating the coefficient of x² and constant term, A+B=0

4+B=0

B=-4

A-C=5

4-C=5

C=-1

From equation (1),we get,

Let

Put t=x²

dt=2xdx

Now putting,

A(t+3) +B(t+1) = 1

Putting t+3=0,

X=-3

A(0) +B(-3+1)=1

Putting t+1=0,

X=-1

A(-1+3)+B(0)=1

From equation(1),we get,

Let

Put t=x²

dt=2xdx

Now putting,

A(t+1)+B(t-1) = t

Putting t+1=0,

t=-1

A(0)+B(-1-1)=-1

Putting t-1=0,

t=1

A(1+1)+B(0)=1

From equation(1),we get,

Let

Put

A(x²+x+1)+(Bx+C)(x-1)=1

Now putting x-1=0

X=1

A(1+1+1)+0=1

By equating the coefficient of x² and constant term, A+B=0

A-C=1

From the equation(1), we get,

Put t=x²+x+1

dt=(2x+1)dx

Let

Put

A(x²-x+1)+(Bx+C)(x+1)=1

Now putting x+1=0

X=-1

A(1+1+1)+C(0)=1

By equating the coefficient of x² and constant term, A+B=0

A+C+=1

From the equation(1), we get,

Let

Put

A(x+1)(x²+1)+B(x²+1)+(Cx+D)(x+1)²=1

Put x+1=0

X=-1

A(0)+B(1+1)+0=1

By equating the coefficient of x² and constant term, A+C=0

A+B+2C=0……(2)

A+B+D=1

Solving (2) and (3),we get,

Let

Put

A(x²+4)+(Bx+C)(2x+1) =17

Put 2x+1=0

A=4

By equating the coefficient of x² and constant term,

A+2B=0

4+2B=0

B=-2

4A+C=17

4×4+C=17

C=1

From the equation(1), we get,

Let

Put

A(t+4)+B(t+2) = 1

Put t+4=0

t=-4

A(0)+B(-4+2)=1

Put t+2=0

t=-2

A(-2+4)+B(0)=1

From equation(1),we get,

Let

Putting

Where t=x²

(A+B)t+(25A+4B)=t+1

A+B=1………….(1)

25A+4B=1………..(2)

Solving equation (1)and(2), we get,

Now,

putting t=e^x-1

e^x=t+1

dt= e^x dx

Put

A(t²)+(Bt+C)(t+1)=1

Put t+1=0

t=-1

A=1

Equating coefficients

A+B=0

1+B=0

B=-1

C=1

From equation (1),we get,

Let

Put t=x⁵

dt=5x⁴dx

Putting

A(t+1)+Bt=1

Now put t+1=0

t=-1

A(0)+B(-1)=1

B=-1

Now put t=0

A(0+1)+B(0)=1

A=1

Let

Put t=x⁶

dt=6x⁵dx

Putting

A(t+1)+Bt=1

Now put t+1=0

t=-1

A(0)+B(-1)=1

B=-1

Now put t=0

A(0+1)+B(0)=1

A=1

let

Put t=cosx

dt=-sinxdx

Putting

A(1+t)(3+2t)+B(1-t)(3+2t)+C(1+t)(1-t)=1

Now Putting 1+t=0

t=-1

A(0)+B(2)(3-2)+C(0)=1

Now Putting 1-t=0

t=1

A(2)(5)+B(0)+C(0)=1

Now Putting 3+2t=0

let

Put t=sinx

dt=cosxdx

Putting

A(1+t)(5-4t)+B(1-t)(5-4t)+C(1+t)(1-t)=1

Now Putting 1+t=0

t=-1

A(0)+B(2)(9)+C(0)=1

Now Putting 1-t=0

t=1

A(2) +B(0)+C(0)=1

Now Putting 5-4t=0

From equation(1),we get,

Let

let dx = dx

Put t=sinx

dt=cosxdx

Putting

A(1+t)² +B(1-t)(1+t)+C(1+t)=t

Now Putting 1-t=0

t=1

A(0)+B(0)+C(1+1)=1

Now Putting 1+t=0

t=-1

A(2)² +B(0)+C(0)=-1

By equating the coefficient of t²,we get,A-B=0

From equation(1),we get,

let =

Put t=cosx

dt=-sinxdx

Putting

A(1+t)(1+2t)+B(1-t)(1+2t)+C(1-t²)=1

Putting 1+t=0

t=-1

A(0)+B(2)(1-2)+C(0)=1

Putting 1-t=0

t=1

A(2)(3)+B(0)+C(0)=1

Putting 1+2t=0

From equation(1),we get,

Let

Putting

Where t=x²

A(t+3)+B(t-4)=t

Now put t+3=0

t=-3

A(0)+B(-7)=-3

Now put t-4=0

t=4

A(4+3)+B(0)=4

From equation(1)

Let

Putting

Where t=x²

t²=A(t+9)(t+16)+B(t+1)(t+16)+C(t+1)(t+9)

Now put t+1=0

t=-1

A(8)(15)+B(0)+C(0)=1

Now put t+9=0

t=-9

A(-9+9)(-9+16)+B(-9+1)(-9+16)+C(-9+1)(-9+9)=(-9)²

A(0)+B(-56)+C(0)=81

Now put t+16=0

t=-16

A(0)+B(0)+C(-15)(-7)=(-16)²

A(0)+B(0)+C(105)=256

From equation(1)

let

Put t=cos2x

dt=-2sin2xdx

Putting

A(1-t)+B(t-2)=1

Putting 1-t=0

t=1

A(0)+B(1-2) =1

B=-1

Putting t-2=0

t=2

A(1-2)+B(0) =1

A=-1

From equation (1), we get,

Let

Put

A(1+x²)+Bx(1-x)+C(1-x) =2

Put x=1

2=2A+0+0

A=1

Put x=0

2=A+C

C=2-A

C=2-1=1

Putting x=2

We have 2=5A-2B-C

2=5×1-2B-1

2B=2

B=1

Let

Again let x²=t

2t+1=A(t+4)+B(t)

Putting t=-4

2(-4)+1=A(-4+4)+B(-4)

-8+1=0-4B

-7=-4B

Putting t=0

2(0)+1=A(0+4)+B(0)

1=4A

Let

Let sin x=t

cos x dx=dt

Let (sin ∅-2)= t

cos ∅ d∅=dt

{1-cos 2x=2 sin² x}

Let log x = t

Let

Let +cosx=t

-sin x dx=dt

Let cos x-sin x=t

-(sin x + cos x)dx=dt

Let sin x + cos x=t

(cos x-sin x)dx=dt

Let (x + log (sin x ))=t

(1+cot x) dx=dt

Let (x + cos² x)=t

(1-sin 2x) dx=dt

Let

Let

Let 1-tan² x=t

-2 tan x. sec² x dx=dt

Using Integration by Parts

Here 1 is the first function and is second function

{2 cos A cos B=cos(A+B)+cos(A-B)}

{ 2 sin A sin B=cos(A-B)-cos(A+B) }

{

Here f(x)=-log cos x

∫e^x (tan x- log cos x )dx=-e^x (log cos x)+c

Multiplying Num^r and Den^r with (1+sinx)

= tan x + sec x + c

Let x² =t

2xdx=dt

Let sin x=t

cos x dx=dt

On rationalizing

We know that,

.

Let (x + log (sec x ))=t

(1+tan x) dx=dt

Let

Let x²+x+1=t

(2x+1)dx=dt

We know that,

We know that,

We know that,

We know that,

We know that,

We know that,

We know,

4x=t

4dx=dt

We know,

put t=4x

2x=t

2dx=dt

We know,

put t=2x

cos x=t

-sin x dx=dt

We know,

put t=cos x

=-tan^-1 (cos x)+c

sin x=t

cos x dx=dt

We know,

put t=sin x

= tan^-1 (sin x)+c

e^x =t

e^x dx=dt

We know,

put t=e^x

tan^-1 e^x +c

Let x⁶ =t

6x⁵ dx=dt

We know,

put t=x⁶

Let x ⁴=t

4x³ dx=dt

We know,

put t=x⁴

Completing the square

x² +4x+8 = x²+4x+8 (+4-4)

=x²+4x+4+4

=(x+2)² +2²

Let x+2=t

dx=dt

We know,

put t=x+2

Completing the square

dx=dt

We know,

e^x =t e^x

e^x dx=dt

We know,

put t= e^x

= tan^-1 e^x +c

Let 2x=t

2 dx=dt

We know,

put t=2x

Consider,

3x=t

3dx=dt

We know,

put t=3x

Consider ,

Completing the square

dx=dt

We know,

put t=x- �

tan x=t

sec² x dx=dt

Let t-t^-1 =u

1+x^-2 dt=du

We know,

put u=t-t^-1

put t=tan x

Let x-x^-1 =t

1+x^-2 dx=dt

We know,

put t=x-x^-1

Let sec² x-1=t

2 sec x sec x tan x dx=dt

We know,

=tan^-1 t +c

put t=sec² x-1

= tan^-1 sec²x – 1 + c

= tan^-1 tan²x + c

Consider

3x=t

3dx=dt

We know,

put t=3x

Consider

2x=t

2dx=dt

We know,

put t=2x

Let x³ =t

3x² dx=dt

We know,

put t=x³

Let x² =t

2x dx=dt

We know,

put t=x²

Let x³ =t

3x² dx=dt

We know,

put t=x³

Completing the square

x² +2x-3 = x² +2x-3+1-1

(x+1)²-4

Let x+1=t

dx=dt

We know,

put t=x+1

Let √3 tan x=t

√3 sec² x dx=dt

We know,

put t=√3 tan x

Let cot x=t

-cosec² x dx=dt

We know,

put t=cot x

Consider

2x=t

2dx=dt

We know,

put t=2x

Let x² =t

2x dx=dt

We know,

put t=x²

Let tan x=t

sec² x dx=dt

We know,

put t=tan x

Let 2 tan x=t

2 sec² x dx=dt

We know,

put t=2 tan x

Let tan x=t

sec² x dx=dt

We know,

put t= tan x

Let x-x^-1=t

1+x^-2 dx=dt

We know,

put t=x-x^-1

Put sin x =t

⇒ cos x dx = dt

∴ The given equation becomes

But t = sin x

⇒ Let t=2^x

dt = log 2. 2^x.dx

But t =2^x

=sin^-1 (x-1)+c

⇒

Let x³ =t

∴ 3x²dx = dt

∴ x⁶ =t²

⇒

⇒

But t =x³

⇒

⇒

⇒

⇒

⇒

= sin^-1(2x-1)+c

We know

Put t =2x

dt =2 dx

⇒

But t = 2x

⇒

Put t =x³

dt =3x²dx

But t = x³

⇒

Put t =2cos x

dt =-2sinxdx

But t = 2cosx

⇒

Put t =tanx

dt = sec²x

But t = tanx

Differentiating both side with respect to t

Again put, t = 1 – e^2x

.

⇒

Let t =sin x

dt =cos x dx

But t =sin x

Put t =x³

dt =3x²dx

But t = x³

Put t =tan x

dt = sec²x

But t = tan x

Let x³ = t

⇒ 3x²dx = dt

But t =x³

We know

We know

We know

Given:

Let sin x =t

cos x dx =dt

But t =sin x

We know

It can be written as

We know

Let e^x = t

e^x dx =dt

But t =e^x

Let log x =t

But t =log x