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Question Number 123921 by john_santu last updated on 29/Nov/20

$$\int \frac{1}{\sqrt{x}(x+1)({\left({\tan }^{-1}\sqrt{x}\right)}^2+9)}dx$$

Answered by mindispower last updated on 29/Nov/20

$$=\int \frac{2}{(t^2+1)({tan}^{-}{\left(t\right)}^2+9)}$$$$=\int \frac{2}{(1+t^2)(1+{\left(\frac{{tan}^{-}\left(t\right)}{3}\right)}^2)}dt$$$$=\int \frac{6d\left(\frac{{tan}^{-}\left(t\right)}{3}\right)}{(1+{\left(\frac{{tan}^{-}\left(\mathit{t}\right)}{3}\right)}^2)}=6{\mathit{t}\mathit{a}\mathit{n}}^{-}\left(\frac{{\mathit{t}\mathit{a}\mathit{n}}^{-}\left(\mathit{t}\right)}{3}\right)$$$$\mathit{t}=\sqrt{\mathit{x}}$$$$\mathit{w}\mathit{e}\mathit{g}\mathit{e}\mathit{t}6{\mathit{t}\mathit{a}\mathit{n}}^{-}\left(\frac{{\mathit{t}\mathit{a}\mathit{n}}^{-}\left(\sqrt{\mathit{x}}\right)}{3}\right)+\mathit{c}$$

Commented by Mammadli last updated on 29/Nov/20

Commented by liberty last updated on 29/Nov/20

$$=\frac{1}{5}\int \sec {}^2\left(5x\right)d\left(5x\right)=\frac{1}{5}\tan \left(5x\right)+c$$$$\underset{\pi /6}{\stackrel{\pi /4}{\int }}\frac{dx}{\cos {}^2\left(5x\right)}=\frac{1}{5}(\tan \frac{5\pi }{4}-\tan \frac{5\pi }{6})$$

Commented by Dwaipayan Shikari last updated on 29/Nov/20

$${\int }_{\frac{\pi }{6}}^{\frac{\pi }{4}}{sec}^{2}5xdx=\frac{1}{5}{\int }_{\frac{5\pi }{6}}^{\frac{5\pi }{4}}{sec}^{2}udu=\frac{1}{5}{\left[tanu\right]}_{\frac{5\pi }{6}}^{\frac{5\pi }{4}}=\frac{\sqrt{3}-1}{5\sqrt{3}}$$

Answered by Olaf last updated on 29/Nov/20

$$\mathrm{Let}\sqrt{x}=\tan \theta$$$$\int \frac{2\tan \theta (1+{\tan }^2\theta )d\theta }{\tan \theta ({\tan }^2\theta +1)({\theta }^2+9)}$$$$2\int \frac{d\theta }{{\theta }^2+9}=2\times \frac{1}{3}\arctan \left(\frac{\theta }{3}\right)=\frac{2}{3}\arctan \left(\frac{\arctan \sqrt{x}}{3}\right)$$

Answered by liberty last updated on 29/Nov/20

$$lettingz={\tan }^{-1}\sqrt{x}\Rightarrow \sqrt{x}=\tan z$$$$\frac{dx}{2\sqrt{x}}=\sec {}^{2}zdz$$$$Y=\int \frac{2\sec {}^{2}zdz}{(\tan {}^{2}z+1)(z^2+9)}$$$$Y=\int \frac{2dz}{z^2+3^2}=\frac{2}{3}\arctan \left(\frac{z}{3}\right)+c$$$$Y=\frac{2}{3}\arctan \left(\frac{{\tan }^{-1}\sqrt{x}}{3}\right)+c$$

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