UPSC NDA Math Model Paper 3

Here we have to find the value of ∫

x

(x+2)(3−2x)

dx
⇒

x

(x+2)(3−2x)

=

A

(x+2)

+

B

(3−2x)

⇒x=A(3−2x)+B(x+2) ........(1)
By putting x = - 2 on both the sides of (1) we get A = - 2/7
By putting x = 3/2 on both the sides of (1) we get B = 3/7
⇒∫

x

(x+2)(3−2x)

dx=−

2

7

‌∫

dx

x+2

+

3

7

‌∫

dx

3−2x

As we know that ∫

dx

x

=log|x|+C where C is a constant
⇒∫

x

(x+2)(3−2x)

dx =−

2

7

‌log|x+2|−

3

14

‌log|3−2x|+C where C is a constant