int frac{d x}{left(2 a x+x^2right)^{frac{3}{2 | vedclass

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(D) Let $I = \int \frac{d x}{\left(2 a x+x^2\right)^{3 / 2}} = \int \frac{d x}{\left((x+a)^2-a^2\right)^{3 / 2}}$.Substitute $t = x+a$,then $d t = d x

Vedclass · Accepted Answer

$\frac{-1}{a^2} \left( \frac{x+a}{\sqrt{2 a x+x^2}} \right) + C$

Vedclass · Answer

(D) Let $I = \int \frac{d x}{\left(2 a x+x^2ight)^{3 / 2}} = \int \frac{d x}{\left((x+a)^2-a^2ight)^{3 / 2}}$.Substitute $t = x+a$,then $d t = d x$.$I = \int \frac{d t}{\left(t^2-a^2ight)^{3 / 2}}$.Let $t = a \sec 	heta$,then $d t = a \sec 	heta 	an 	heta \, d 	heta$.$I = \int \frac{a \sec 	heta 	an 	heta}{\left(a^2 \sec^2 	heta - a^2ight)^{3 / 2}} \, d 	heta = \int \frac{a \sec 	heta 	an 	heta}{\left(a^2 	an^2 	hetaight)^{3 / 2}} \, d 	heta$.$I = \int \frac{a \sec 	heta 	an 	heta}{a^3 	an^3 	heta} \, d 	heta = \frac{1}{a^2} \int \frac{\sec 	heta}{	an^2 	heta} \, d 	heta = \frac{1}{a^2} \int \frac{\cos 	heta}{\sin^2 	heta} \, d 	heta$.$I = \frac{1}{a^2} \int \csc 	heta \cot 	heta \, d 	heta = -\frac{1}{a^2} \csc 	heta + C$.Since $\sec 	heta = \frac{t}{a}$,we have $\cos 	heta = \frac{a}{t}$,so $\sin 	heta = \sqrt{1 - \frac{a^2}{t^2}} = \frac{\sqrt{t^2-a^2}}{t}$.Thus,$\csc 	heta = \frac{t}{\sqrt{t^2-a^2}}$.Substituting back,$I = -\frac{1}{a^2} \left( \frac{t}{\sqrt{t^2-a^2}} ight) + C = -\frac{1}{a^2} \left( \frac{x+a}{\sqrt{(x+a)^2-a^2}} ight) + C$.$I = -\frac{1}{a^2} \left( \frac{x+a}{\sqrt{2 a x+x^2}} ight) + C$.

$\int \frac{d x}{\left(2 a x+x^2\right)^{\frac{3}{2}}} = $

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