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Question Number 12742 by Joel577 last updated on 30/Apr/17

Prove that

\int \frac{d x}{{(x + 1)}^{2} \sqrt{x^{2} + 2 x + 2}} = \frac{- \sqrt{x^{2} + 2 x + 2}}{x + 1} + C

Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 30/Apr/17

\frac{1}{x + 1} = t \Rightarrow \frac{d x}{{(x + 1)}^{2}} = - d t

I = \int \frac{- d t}{\sqrt{\frac{1}{t^{2}} + 1}} = \int \frac{- t d t}{\sqrt{t^{2} + 1}} = - \sqrt{t^{2} + 1} + C =

- \sqrt{\frac{1}{{(x + 1)}^{2}} + 1} + C = - \frac{\sqrt{x^{2} + 2 x + 2}}{x + 1} + C .