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# Integrate : $\int \sqrt{\large\frac{x}{1+x}}$$dx (a)\;\frac{2}{3}\sqrt{1+x^2}. \cos x+c \\(b)\;\frac{2}{3}\sqrt{1+x^2}. x+c \\(c)\;\frac{2}{3}\sqrt{1-x^2}. \cos x+c\\ (d)\;None Can you answer this question? ## 1 Answer 0 votes Put x= \tan ^2 \theta differentiate with respect to x dx= 2 \tan \theta \sec^2 \theta d \theta => \int \large\frac{\tan \theta }{\sec \theta } \times$$ 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.
edited Mar 20, 2014