• # question_answer Evaluate $\int_{3}^{4}{\frac{\sqrt{x}}{\sqrt{x}+\sqrt{7-x}}dx.}$ OR Evaluate $\int_{0}^{1}{\frac{\log (1+x)}{1+{{x}^{2}}}\,dx.}$

 Let $l=\int\limits_{3}^{4}{\frac{\sqrt{x}}{\sqrt{x}+\sqrt{7-x}}\,\,dx}$                                 ?(i) Using $\int\limits_{a}^{b}{f(x)dx=}\int\limits_{a}^{b}{f(a+b-x)dx}$ in Eq. (i), we get $l=\int\limits_{3}^{4}{\frac{\sqrt{3+4-x}}{\sqrt{3+4-x}+\sqrt{7-(3+4-x)}}dx}$ $=\int\limits_{3}^{4}{\frac{\sqrt{7-x}}{\sqrt{7-x}+\sqrt{x}}}$ On adding Eqs. (i) and (ii), we get $2l=\int\limits_{3}^{4}{\frac{\sqrt{x}}{\sqrt{x}+\sqrt{7-x}}dx+\int\limits_{3}^{4}{\frac{\sqrt{7-x}}{\sqrt{7-x}+\sqrt{x}}dx}}$ $=\int\limits_{3}^{4}{\frac{\sqrt{x}+\sqrt{7-x}}{\sqrt{x}+\sqrt{7-x}}dx+\int\limits_{3}^{4}{dx=[x]_{3}^{4}}}$ $\Rightarrow$   $2l=[4-3]=1$ $\therefore$      $l=\frac{1}{2}$ OR Let $l=\int\limits_{0}^{1}{\frac{\log \,(1+x)}{1+{{x}^{2}}}\,\,dx}$ Put $t={{\tan }^{-1}}x\Rightarrow dt=\frac{1}{1+{{x}^{2}}}dx$ $\therefore$Also, when $x=0,$ then $t=0$ and when $x=1,$ then $t=\frac{\pi }{4}$ $\therefore$      $l=\int\limits_{0}^{\pi /4}{\log \,(tant\,+1)dt=\int\limits_{0}^{\pi /4}{\log \left( \frac{\sin t}{\cos t}+1 \right)dt}}$ $=\int\limits_{0}^{\pi /4}{\log \left( \frac{\sin t+\cos t}{\cos t} \right)}\,\,dt$ $=\int\limits_{0}^{\pi /4}{[log\,(sint+cost)-\log \,(cost)]dt}$ Here, $\sin t+\cos t=\sqrt{2}\left( \frac{1}{\sqrt{2}}\sin t+\frac{1}{\sqrt{2}}\cos t \right)$ $=\sqrt{2}\left[ \sin \frac{\pi }{4}\cdot \sin t+\cos \frac{\pi }{4}\cdot \cos t \right]$ $=\sqrt{2}\cos \left( \frac{\pi }{4}-t \right)$ Now,     $l=\int\limits_{0}^{\pi /4}{\left[ \log \left[ \sqrt{2}\cos \left( \frac{\pi }{4}-t \right) \right]-\log (cost) \right]dt}$ $=\int\limits_{0}^{\pi /4}{\log \sqrt{2}dt+\int\limits_{0}^{\pi /4}{\log \left( \cos \left( \frac{\pi }{4}-t \right) \right)dt}}$ $-\int\limits_{0}^{\pi /4}{\log \,(cos\,t)dt}$ Again, put $u=\frac{\pi }{4}-t\Rightarrow du=-dt$ Also, when $t=0,$then $u=\frac{\pi }{4}$and when $t=\frac{\pi }{4}$then $u=0$ $\therefore$$l=\log \sqrt{2}[t]_{0}^{\pi /4}+\int\limits_{0}^{\pi /4}{\log \,(\cos u)\,du-\int\limits_{0}^{\pi /4}{\log \,(\cos t)\,dt}}$ $=\log \sqrt{2}\left[ \frac{\pi }{4}-0 \right]$ $=\frac{\pi }{8}\log 2$             $[\because \log {{a}^{b}}=b\log a]$