Question Number 20257 by tammi last updated on 24/Aug/17
$$\int\frac{{dx}}{\left(\mathrm{1}−{x}\right)\sqrt{\mathrm{1}+{x}}} \\ $$
Answered by Tinkutara last updated on 24/Aug/17
$${t}^{\mathrm{2}} \:=\:{x}\:+\:\mathrm{1} \\ $$$${dx}\:=\:\mathrm{2}{tdt} \\ $$$$\int\frac{\mathrm{2}{tdt}}{\left(\mathrm{2}\:−\:{t}^{\mathrm{2}} \right){t}}\:=\:−\mathrm{2}\int\frac{{dt}}{{t}^{\mathrm{2}} \:−\:\left(\sqrt{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$$=\:−\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\right)\:\mathrm{log}\:\mid\frac{{t}\:−\:\sqrt{\mathrm{2}}}{{t}\:+\:\sqrt{\mathrm{2}}}\mid\:+\:{C} \\ $$$$=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\mathrm{log}\:\mid\frac{{t}\:+\:\sqrt{\mathrm{2}}}{{t}\:−\:\sqrt{\mathrm{2}}}\mid\:+\:{C} \\ $$$$=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\mathrm{log}\:\mid\frac{\sqrt{{x}\:+\:\mathrm{1}}\:+\:\sqrt{\mathrm{2}}}{\:\sqrt{{x}\:+\:\mathrm{1}}\:−\:\sqrt{\mathrm{2}}}\mid\:+\:{C} \\ $$
Commented by tammi last updated on 25/Aug/17
$${thank}\:{you} \\ $$