# Intergate : $\int \sqrt {\large\frac{x+a}{x}}$$dx (a)\;\sqrt x \sqrt {x-a}+a \log (\sqrt {x+a}+\sqrt x )+c \qquad(b)\;\sqrt x +a \log (\sqrt {x+a}+\sqrt x )+c \qquad(c)\;\sqrt {x+a}+a \log (\sqrt {x+a}+\sqrt x )+c \qquad (d)\;\sqrt x \sqrt {x-a}+a \log (\sqrt {x+a}-\sqrt x )+c ## 1 Answer Need homework help? Click here. Put x= a \tan ^2 \theta dx= 2a \tan \theta . \sec^2 \theta . d \theta \qquad= \int \large\frac{\sec \theta}{\tan \theta}$$. 2a \tan \theta \sec^2 \theta . d\theta$
$\qquad= \int 2a \sec^3 \theta.d\theta$
=> $2a \int \sec^2 \theta. d\theta$
=> $\int \sec^2 \theta . d\theta$
=> $2a [ \sec \theta \tan \theta -\int \sec \theta (\sec^2 \theta-1) d \theta$
=> $2a [\sec \theta \tan \theta -\int (\sec ^3 \theta+ \sec \theta) d\theta]$
=> $\large\frac{2a}{2}$$\sec \theta \tan \theta + \large\frac{2a}{2}$$ \log (\sec \theta+ \tan \theta)+c$
=> $a \sec \theta \tan \theta + a \log /sec \theta + \tan \theta +c$
Put values of $\sec \theta +\tan \theta$
$\sqrt x \sqrt {x-a}+a \log (\sqrt {x+a}+\sqrt x )+c$
Hence a is the correct answer.
answered Dec 24, 2013 by