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∫ 1 √ Tan − 1 X . ( 1 + X 2 ) D X - Mathematics

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Question

\[\int\frac{1}{\sqrt{\tan^{- 1} x} . \left( 1 + x^2 \right)} dx\]
Sum
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Solution

\[\int\frac{dx}{\sqrt{\tan^{- 1} x} \left( 1 + x^2 \right)}\]
\[Let \tan^{- 1} x = t\]
\[ \Rightarrow \frac{1}{1 + x^2} = \frac{dt}{dx}\]
\[ \Rightarrow \frac{1}{1 + x^2}dx = dt\]

\[Now, \int\frac{dx}{\sqrt{\tan^{- 1} x} \left( 1 + x^2 \right)}\]


\[ = \int \frac{dt}{\sqrt{t}}\]
\[ = \int t^{- \frac{1}{2}} dt\]


\[ = \frac{t^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1} + C\]


\[ = 2 \sqrt{t} + C\]
\[ = 2 \sqrt{\tan^{- 1} x} + C\]

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Chapter 19: Indefinite Integrals - Exercise 19.09 [Page 58]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 19 Indefinite Integrals
Exercise 19.09 | Q 15 | Page 58

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