# Integrate: $\int \large\frac{2}{3} \frac{e^x \sqrt {e^x-4}}{e^x+g}$$dt $\begin {array} {1 1} (a)\;\frac{6 \sqrt 2}{5}-26 \tan ^{-1} \bigg(\sqrt {\frac{8}{13}}\bigg) \\ (b)\;\frac{8 \sqrt 2}{3}-\frac{52}{3} \tan ^{-1} \bigg(\sqrt {\frac{8}{13}}\bigg) \\ (c)\;\frac{2 \sqrt 2}{4}-39 \tan ^{-1} \bigg(\sqrt {\frac{8}{13}}\bigg) \\ (d)\;\frac{8 \sqrt 2}{5}-\frac{52}{3} \sin ^{-1} \bigg(\sqrt {\frac{8}{13}} \bigg) \end {array}$ ## 1 Answer e^x-2=t^2 e^x.dx= 2t \;dt x= n4 lower limit =0 x=ln12 upper limit =\sqrt {\theta} => \large\frac{2}{3} \int \limits_0^{\sqrt {\theta}} \frac{2t.t}{t^2+13}$$dt$
=> $\large\frac{2}{3}$$\bigg[2 \int \limits_0^{\sqrt {\theta}} \large\frac{t^2+13}{t^2+13}$$dt-2 \int \limits_0^{\sqrt {\theta}} \large\frac{13}{t^2+13}dt\bigg]$
=>$\large\frac{2}{3}$$\bigg[[2t]_0^{\sqrt {\theta}}-2.13.\bigg[\tan^{-1}(\frac{t}{\sqrt{13}})\bigg]_0^{\sqrt {\theta}} \bigg] => \large\frac{2}{3}$$[ 2 \sqrt {\theta}-0-2 .12.\tan ^{-1} \bigg(\sqrt {\large\frac{\theta}{13}}\bigg)-2.13. \tan^{-1}(\theta)\bigg]$
=> $\large\frac{2}{3}$$\bigg[4 \sqrt 2 - 26 \tan^{-1} \bigg(\sqrt {\large\frac{\theta}{13}}\bigg)\bigg] \large\frac{8 \sqrt 2}{3}-\frac{52}{3}$$ \tan ^{-1} \bigg(\sqrt {\frac{8}{13}}\bigg)$
Hence b is the correct answer.