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Answer : -        $\frac{1}{\sqrt{3}} \tan ^{-1}\left(\sqrt{\frac{x^{2}+1}{3}}\right)+C$

Hint :-                Use substitution method and special integration formula..

Given :-                $\int \frac{x}{\left(x^{2}+4\right) \sqrt{x^{2}+1}} d x$

Sol : -             Let $\mathrm{I}=\int \frac{x}{\left(x^{2}+4\right) \sqrt{x^{2}+1}} d x$

$\begin{array}{r} \text { Put } x^{2}+1=t^{2} \Rightarrow 2 \mathrm{xd} \mathrm{x}=2 \mathrm{t} \mathrm{dt} \rightarrow \mathrm{x} \mathrm{dx}=\mathrm{t} \mathrm{dt} \text { than, } \\\\ \mathrm{I}=\int \frac{1}{\left(t^{2}+3\right) \sqrt{t^{2}}} \mathrm{t} \mathrm{dt}\left(\because x^{2}+1=\mathrm{t}^{2} \rightarrow x^{2}=t^{2}-1\right. \\\\ \left.\rightarrow x^{2}+4=t^{2}-1+4 \rightarrow x^{2}+4=t^{2}+3\right) \end{array}$

\begin{aligned} &\mathrm{I}=\int \frac{1}{\left(t^{2}+3\right) \cdot t} \mathrm{t} \mathrm{dt} \\ &\mathrm{I}=\int \frac{1}{t^{2}+3} \mathrm{dt} \\ &\mathrm{I}=\int \frac{1}{t^{2}+(\sqrt{3})^{2}} \mathrm{dt} \end{aligned}

\begin{aligned} &I=\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{t}{\sqrt{2}}\right)+c \quad\left[\because \int \frac{1}{a^{2}+x^{2}} d x=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+c\right] \\ &I=\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{\sqrt{x^{2}+1}}{\sqrt{3}}\right)+c \\ &I=\frac{1}{\sqrt{3}} \tan ^{-1}\left(\sqrt{\frac{x^{2}+1}{3}}\right)+c \quad\left[\because t=\sqrt{\left.x^{2}+1\right]}\right. \end{aligned}

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