again posting on behalf of vc15ao4
i assume you mean (t^2 + 1)/(t + 1), since any alternate translation would be too easy.Originally Posted by vc15ao4
int{(t^2 + 1)/(t + 1)}dt
let u = t + 1
=> du = dt
if u = t+1
=> t = u - 1
=> t^2 = (u - 1)^2
=> t^2 + 1 = (u - 1)^2 + 1 = u^2 - 2u + 2
so our integral becomes:
int{(u^2 - 2u + 2)/u}du
= int{u - 2 + 2/u}du
= (1/2)u^2 - 2u + 2ln(u) + C
= (1/2)(t + 1)^2 - 2(t + 1) + 2ln(t + 1) + C
= (1/2)(t + 1)[(t + 1) - 4] + 2ln(t + 1) + C
= (1/2)(t + 1)(t - 3) + 2ln(t + 1) + C