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

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Question

(A) Let $I = \int \frac{dx}{(2ax+x^2)^{3/2}}$.We can rewrite the expression inside the integral as:$I = \int \frac{dx}{(x^2+2ax)^{3/2}} = \int \frac{d

Vedclass · Accepted Answer

$\frac{-1}{a^2} \frac{(x+a)}{\sqrt{2 a x+x^2}}+c$

Vedclass · Answer

(A) Let $I = \int \frac{dx}{(2ax+x^2)^{3/2}}$.We can rewrite the expression inside the integral as:$I = \int \frac{dx}{(x^2+2ax)^{3/2}} = \int \frac{dx}{[x(x+2a)]^{3/2}} = \int \frac{dx}{x^{3/2}(x+2a)^{3/2}}$.Alternatively, complete the square inside the bracket:$2ax+x^2 = (x+a)^2 - a^2$.Let $x+a = a \sec 	heta$, then $dx = a \sec 	heta 	an 	heta d	heta$.Also, $(x+a)^2 - a^2 = a^2 \sec^2 	heta - a^2 = a^2 	an^2 	heta$.Substituting these into the integral:$I = \int \frac{a \sec 	heta 	an 	heta d	heta}{(a^2 	an^2 	heta)^{3/2}} = \int \frac{a \sec 	heta 	an 	heta d	heta}{a^3 	an^3 	heta} = \frac{1}{a^2} \int \frac{\sec 	heta}{	an^2 	heta} d	heta$.$I = \frac{1}{a^2} \int \frac{1/\cos 	heta}{\sin^2 	heta / \cos^2 	heta} d	heta = \frac{1}{a^2} \int \frac{\cos 	heta}{\sin^2 	heta} d	heta$.Let $u = \sin 	heta$, then $du = \cos 	heta d	heta$.$I = \frac{1}{a^2} \int u^{-2} du = \frac{1}{a^2} (-u^{-1}) + c = -\frac{1}{a^2 \sin 	heta} + c$.Since $\sec 	heta = \frac{x+a}{a}$, we have $\cos 	heta = \frac{a}{x+a}$.Then $\sin 	heta = \sqrt{1 - \cos^2 	heta} = \sqrt{1 - \frac{a^2}{(x+a)^2}} = \sqrt{\frac{(x+a)^2 - a^2}{(x+a)^2}} = \frac{\sqrt{x^2+2ax}}{x+a}$.Therefore, $I = -\frac{1}{a^2} \cdot \frac{x+a}{\sqrt{x^2+2ax}} + c$.

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

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