Question

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

Answer 1

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

Let 1 + x² = t

2xdx = dt

`xdx = dt/2`

= `1/2 int 1/((t - 1)t) dt`

`1/((t - 1)(t)) = A/t + B/(t - 1)`   ...(i)

1 = A(t – 1) + Bt

Put t = 0

A = –1

Put t = 1

B = 1

Put A = –1 and B = 1   ...(ii)

`1/((t - 1)t) = (-1)/t + 1/(t - 1)`

`1/2 int 1/((t - 1)t) dt = 1/2 [-int 1/t dt + int 1/(t - 1) dt]`

= `1/2 [-log t + log |t - 1|] + C`

= `1/2 [log  (t - 1)/t] + C`

Put t = x² + 1

= `1/2 [log  (x^2 + 1 - 1)/(x^2 + 1)] + C`

= `1/2 log  x^2/(1 + x^2) + C`