Evaluate : e^ ^ -1 x (1+x+x^2 1+x^2 ) d x - Vidyadip

Question

Evaluate : $\int e^{\tan ^{-1} x} \cdot\left(\frac{1+x+x^2}{1+x^2}\right) \cdot d x$

✓

Answer

put $\tan ^{-1} x=t$
$
\therefore x=\tan t
$
differentiating $w . r . t . x$
$
\begin{aligned}
& \therefore \frac{1}{1+x^2} \cdot d x=1 \cdot d t \\
& \mathrm{I}=\int e^t \cdot\left[1+\tan t+\tan ^2 t\right] \cdot d t \\
& =\int e^t \cdot\left[\tan t+\left(1+\tan ^2 t\right)\right] \cdot d t \\
& =\int e^t \cdot\left[\tan t+\sec ^2 t\right] \cdot d t \\
& \text { Here } f(t)=\tan t \\
& \Rightarrow \quad f^{\prime}(t)=\sec ^2 t \\
& \mathrm{I}=e^t \cdot f(t)+c \\
& =e^t \cdot \tan t+c \\
& =e^{\tan ^{-1} x} \cdot x+c \\
& \therefore \int e^{\tan ^{-1} x} \cdot\left(\frac{1+x+x^2}{1+x^2}\right) \cdot d x=e^{\tan ^{-1} x} \cdot x+c \\
\end{aligned}
$

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