Question Number 12742 by Joel577 last updated on 30/Apr/17
$$\mathrm{Prove}\:\mathrm{that} \\ $$$$\int\:\frac{{dx}}{\left({x}\:+\mathrm{1}\right)^{\mathrm{2}} \:\sqrt{{x}^{\mathrm{2}} \:+\:\mathrm{2}{x}\:+\mathrm{2}}}\:=\:\frac{−\sqrt{{x}^{\mathrm{2}} \:+\:\mathrm{2}{x}\:+\:\mathrm{2}}}{{x}\:+\:\mathrm{1}}\:+\:{C} \\ $$
Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 30/Apr/17
$$\frac{\mathrm{1}}{{x}+\mathrm{1}}={t}\Rightarrow\frac{{dx}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }=−{dt} \\ $$$${I}=\int\frac{−{dt}}{\:\sqrt{\frac{\mathrm{1}}{{t}^{\mathrm{2}} }+\mathrm{1}}}=\int\frac{−{tdt}}{\:\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}}=−\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}+\boldsymbol{{C}}= \\ $$$$−\sqrt{\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }+\mathrm{1}}+\boldsymbol{{C}}=−\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}}}{{x}+\mathrm{1}}+\boldsymbol{{C}}.\:\blacksquare \\ $$