Evaluation of Double integral (4) Qtestion(1) ∫01​∫01+x2​​(1+x2+y2)dxdy​ ..

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Evaluation of Double integral (4) Qtestion(1) Soln \[ \begin{array}{l} \int_{x=0} y=0 \\ =\int_{x=0}^{x=1}\left[\frac{1}{\sqrt{1+x^{2}}} \tan ^{-1}\left(\frac{y}{\sqrt{1+x^{2}}}\right)\right]_{y=0}^{y 0 \sqrt{1+x^{2}}} d x \\ =\int_{x=0}^{x=1} \frac{1}{\sqrt{1+x^{2}}}\left(\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}}{\sqrt{1+x^{2}}}\right)-\tan ^{-1}(\theta)\right) d x \\ =\int_{x=0}^{x=1} \frac{1}{\sqrt{1+x^{2}}}\left(\tan ^{-1}(1)-0\right) d x \\ =\int_{x=0}^{x=1} \frac{1}{\sqrt{1+x^{2}}} \tan ^{-1}(\tan \pi / 4) d x \\ =\pi / 4 \int_{x=0}^{x=1} \frac{1}{\sqrt{1+x^{2}}} d x \\ =\pi / 4\left[\log x+\sqrt{1+x^{2}}\right]_{x=0}^{x=1} \\ =\pi / 4[\log (1+\sqrt{2})-\log (1)] \\ \int_{0}^{1} \cdot \int_{0}^{\sqrt{1+x^{2}} \frac{d x d y}{1+x^{2}+y^{2}}}=\pi / 4 \log (1+\sqrt{2}) \\ \frac{1}{1} x \end{array} \] Scanned with CamScanner

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