Question

solve by partial fraction
integrate

2x/[(x-1)^2(x+2)​]
​

Answer 1

Answer:

P=∫

x(1−2x)

1−x

2

dx

⇒∫

x−2x

2

1−x

2

dx=∫

x(1−2x)

1

dx−∫

x(1−2x)

x

2

dx

⇒∫

x(1−2x)

1

dx−∫

1−2x

x

dx

Let M=∫

x(1−2x)

1

& N=∫

1−2x

x

dx

⇒ using partial fraction in M we get

1=A(1−2x)+Bx

1=x(−2A+B)+A

[A=1] & −2A+B=0

[B=2A=2]

Hence

M=∫

x

1

+

1−2x

2

dx=logx+2log(1−2x) & N=∫

1−2x

x

dx

Let 1−2x=v & x=(1−v)/2

−2dx=dv

N=

4

−1

∫1−v.dv=

4

−1

[v−

2

v

2

]

=

4

−1

[1−2x−

2

(1−2x)

2

]

=−

8

4x

2

+1−4x+4x−2

=

8

4x

2

−1

Hence

P=M+N=log

(1−2x)

2

x

+

8

4x

2

−1

+c

Answer 2

Step-by-step explanation:

2x/[(x-1)^2(x+2)]=A/x+2 +B/(x-1)^2 +C/(x-1)

now u can easily solve it