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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class12  >>  Integral Calculus

Integrate : $\int \sqrt{\large\frac{x}{1+x}}$$dx$

$(a)\;\frac{2}{3}\sqrt{1+x^2}. \cos x+c \\(b)\;\frac{2}{3}\sqrt{1+x^2}. x+c \\(c)\;\frac{2}{3}\sqrt{1-x^2}. \cos x+c\\ (d)\;None$
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1 Answer

Put $x= \tan ^2 \theta $
differentiate with respect to x
$dx= 2 \tan \theta \sec^2 \theta d \theta $
=> $\int \large\frac{\tan \theta }{\sec \theta } \times $$ 2 \tan \theta \sec^2 \theta d \theta$
=> $ 2 \int \tan^2 \theta \sec \theta d \theta$
=> $2 \int \sec^3 \theta d \theta - 2 \int \sec \theta d \theta$
=> $2 . \int \sec \theta . \sec^2 \theta d \theta - 2 \int \sec \theta d \theta $
=> $ 2 . \int \sec \theta \tan \theta - 2 \int \sec \theta \tan^2 \theta d \theta - 2 \int \sec \theta $
=> $ 2 . \int \sec \theta \tan \theta - 2 \int \sec^3 \theta d \theta + 2 \int \sec \theta d \theta$
=> $ 2 . \int \sec \theta \tan \theta - 2 \int \sec^3 \theta d \theta $
=> $\large\frac{2}{3} $$ \sec \theta \tan \theta+c$
$\large\frac{2}{3}$$\sqrt{1+x^2}. x+c$
Hence b is the correct answer.
answered Dec 30, 2013 by meena.p
edited Mar 20, 2014 by balaji.thirumalai
 
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