Question

Find `int (2x)/(x^2 + 1)(x^2 + 2)^2 dx`

Answer 1

`int (2x)/((x^2 + 1)(x^2 + 2)^2) dx`

Let x² = y

⇒ 2xdx = dy

`=> dx = (dy)/(2x)`

`int (2x)/((x^2 + 1)(x^2 + 2)^2)  dx`

= `int (dy)/((y + 1)(y + 2)^2)`

Let `1/((y + 1)(y + 2)^2) = A/(y + 1) + B/(y + 2) + C/(y + 2)^2`  ......(1)

⇒ 1 = A(y+2)² + B(y+1)(y+2) + C(y+1) .....(2)

Putting y = - 2 in (2)

1=C(-2 + 1)

⇒ C = -1

Putting y=−1 in (2)

1 = A(-1 + 2)²

⇒ 1 = A(1)

⇒ A = 1

Putting y=0 in (2)

1 = 4A + B(2) +C

⇒ 1 = 4 + 2B − 1

⇒ 1 = 3 + 2B

⇒ -2 = 2B

⇒ B = -1

Substituting the values of A, B and C in (1)

`1/((y + 1)(y +2)^2) = 1/(y + 1) - 1/(y + 2) - 1/(y + 2)^2`

`=> int (dy)/((y + 1)(y + 2)^2) = int (dy)/(y + 1) - int (dy)/(y + 2) - int (dy)/(y + 2)^2`

= `log |y + 1| - log|y + 2| + 1/(y + 2) + C`

Hence, `int (2x)/((x^2 + 1)(x^2 + 2)^2) dx = log|x^2 + 1|- log |x^2 + 2|+ 1/(x^2 + 2) + C`