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INTEGRALS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsINTEGRALS4 Marks
Question
$\int\text{e}^{\tan^{-1}\text{x}}\Big(\frac{1+\text{x}+\text{x}^2}{1+\text{x}^2}\Big)\text{dx}$
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Answer

Let $\text{I}=\int\text{e}^{\tan^{-1}\text{x}}\Big(\frac{1+\text{x}+\text{x}^2}{1+\text{x}^2}\Big)\text{dx}$
$\int\text{e}^{\tan^{-1}\text{x}}\bigg(\frac{1+\text{x}^2}{1+\text{x}^2}+\frac{\text{x}}{1+\text{x}^2}\bigg)\text{dx}$
$\int\text{e}^{\tan^{-1}\text{x}}\text{dx}+\int\frac{\text{x e}^{\tan^{-1}\text{x}}}{1+\text{x}^2}\text{dx}$
$\text{I}=\text{I}_1+\text{I}_2\ \ \dots(\text{i})$
Now, $\text{I}_2=\int\frac{\text{x e}^{\tan^{-1}\text{x}}}{1+\text{x}^2}\text{dx}$
Put $\tan^{-1}\text{x}=\text{t}\Rightarrow\text{x}=\tan\text{t}$
$\Rightarrow\ \frac{1}{1+\text{x}^2}\text{dx}=\text{dt}$
$\therefore\ \text{I}=\int\tan\text{t}\cdot\text{e}^\text{t}\text{dt}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{I }\ \ \ \ \ \ \ \ \ \text{II}$
$=\tan\text{t}\cdot\text{e}^\text{t}-\int\sec^2\text{t}\cdot\text{e}^\text{t}\text{dt}+\text{C}$
$=\tan\text{t}\cdot\text{e}^\text{t}-\int(1+\tan^2\text{t})\text{e}^\text{t}\text{dt}+\text{C}$ $[\because\sec^2\theta=1+\tan^2\theta]$
$\text{I}_2=\tan\text{t}\cdot\text{e}^\text{t}-\int(1+\text{x}^2)\frac{\text{e}^{\tan^{-1}\text{x}}}{1+\text{x}^2}\text{dx}+\text{C}$
$\text{I}_2=\tan\text{t}\cdot\text{e}^\text{t}-\int\text{e}^{\tan^{-1}\text{x}}\text{dx}+\text{C}$
$\therefore\ \text{I}=\int\text{e}^{\tan^{-1}\text{x}}\text{dx}+\tan\text{t}\cdot\text{e}^\text{t}-\int\text{e}^{\tan^{-1}\text{x}}\text{dx}+\text{C}$
$=\tan\text{t}\cdot\text{e}^\text{t}+\text{C}$
$=\text{x}\text{ e}^{\tan^{-1}\text{x}}+\text{C}$

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