Let
I=∫‌

x+1

x(1−2x)

‌dx
Also, let
‌‌ Also, let ‌‌‌

x+1

x(1−2x)

‌=‌

A

x

+‌

B

1−2x

⇒‌

x+1

x(1−2x)

‌=‌

A(1−2x)+Bx

x(1−2x)

⇒‌x+1‌=x(−2A+B)+A
Comparing coefficient of x both sides, we get
1=−2A+B
Comparing constant term both sides, we get
A=1
∴‌‌ From equation (i),
1=−2(1)+B
⇒‌‌B=3
∴‌‌∫‌

x+1

x(1−2x)

‌dx=∫‌

1

x

‌dx+∫‌

3

1−2x

‌dx
=log|x|+‌

3‌log|1−2x|

−2

+C
=log|x|−‌

3

2

‌log|1−2x|+C‌ Ans. ‌
‌