Integrate the function e^ a ^ -1 x (1+x^2 )^ 3 / 2 with respect t… - Vidyadip

Question

Integrate the function $\frac{e^{a \tan ^{-1} x}}{\left(1+x^2\right)^{3 / 2}}$ with respect to $x$.

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Answer

$
\begin{array}{l}
\text { Let } \\
I=\int \frac{e^{a \tan ^{-1} x}}{\left(1+x^2\right)^{3 / 2}} d x \\
\text { Putting } \quad \tan ^{-1} x=t \\
\therefore \quad x=\tan t \\
\therefore \frac{1}{1+x^2} d x=d t \\
I=\int \frac{e^{a \tan ^{-1} x}}{\sqrt{1+x^2}\left(1+x^2\right)} d x \\
=\int \frac{e^{a t} d t}{\sqrt{1+\tan ^2 t}}=\int \frac{e^{a t} d t}{\sec t} \\
I=\iint_{II}^{a t} \underset{I}{\cos t} d t \\
\text { (using ILATE) } \\
I=\cos t \cdot \frac{e^{a t}}{a}-\int \frac{(-\sin t) \cdot e^{a t}}{a} d t \\
=\frac{1}{a} e^{a t} \cos t+\frac{1}{a} \int_{II} e^{a t} \sin _{I} t d t
\end{array}
$
$\begin{array}{l}=\frac{1}{a} e^{a t} \cos t+ \frac{1}{a}\left(\frac{\sin t \cdot e^{a t}}{a}-\int \cos t \cdot \frac{e^{a t}}{a} d t\right) \\ =\frac{1}{a} e^{a t} \cos t+\frac{1}{a^2} e^{a t} \sin t-\frac{1}{a^2} \int e^{a t} \cos t d t \\ =\frac{1}{a^2} e^{a t}(a \cos t+\sin t)-\frac{1}{a^2} I \\ \left(1+\frac{1}{a^2}\right) I =\frac{e^{a t}}{a^2}(a \cos t+\sin t)+ C \\ I =\frac{e^{a t}}{a^2} \times \frac{a^2}{\left(a^2+1\right)}(a \cos t+\sin t)+ C \\ = \frac{e^{a t}}{\left(a^2+1\right)}(a \cos t+\sin t)+ C \\ \therefore \tan -1 x=t \Rightarrow x=\tan t \\ = \frac{e^{a \tan ^{-1} x}}{a^2+1}\left(a \cdot \frac{1}{\sqrt{1+x^2}}+\frac{x}{\sqrt{1+x^2}}\right)+ C \\ = \frac{e^{a \tan ^{-1} x}}{\left(1+a^2\right)} \cdot \frac{(a+x)}{\sqrt{1+x^2}}+ C \quad Ans .\end{array}$

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