Evaluate the following integrals:
We can write x2 + 2x + 1 + 1 = (x + 1)2 + 1
⇒
Assume x + 1 = tant
⇒ dx = sec2t.dx
⇒ tan2t + 1 = sec2t.
⇒ log|sint| + c
But tant = x + 1
The final answer is