int frac{2x + 1}{(x^2 + 4x + 1)^{3/2}} , dx | vedclass

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(D) Let $I = \int \frac{2x + 1}{(x^2 + 4x + 1)^{3/2}} \, dx$.Rewrite the numerator as $2x + 1 = (2x + 4) - 3$.Then $I = \int \frac{2x + 4}{(x^2 + 4x +

Vedclass · Accepted Answer

$\frac{x-2}{\sqrt{x^2 + 4x + 1}} + C$

Vedclass · Answer

(D) Let $I = \int \frac{2x + 1}{(x^2 + 4x + 1)^{3/2}} \, dx$.Rewrite the numerator as $2x + 1 = (2x + 4) - 3$.Then $I = \int \frac{2x + 4}{(x^2 + 4x + 1)^{3/2}} \, dx - 3 \int \frac{1}{(x^2 + 4x + 1)^{3/2}} \, dx$.For the first part,let $u = x^2 + 4x + 1$,then $du = (2x + 4) \, dx$.$\int u^{-3/2} \, du = -2u^{-1/2} = -\frac{2}{\sqrt{x^2 + 4x + 1}}$.For the second part,complete the square: $x^2 + 4x + 1 = (x + 2)^2 - 3$.Let $x + 2 = \sqrt{3} \sec 	heta$,then $dx = \sqrt{3} \sec 	heta 	an 	heta \, d	heta$.The integral becomes $\int \frac{\sqrt{3} \sec 	heta 	an 	heta \, d	heta}{(3 	an^2 	heta)^{3/2}} = \int \frac{\sqrt{3} \sec 	heta 	an 	heta}{3\sqrt{3} 	an^3 	heta} \, d	heta = \frac{1}{3} \int \frac{\cos 	heta}{\sin^2 	heta} \, d	heta = -\frac{1}{3 \sin 	heta} = -\frac{x+2}{3\sqrt{x^2 + 4x + 1}}$.Combining these,$I = -\frac{2}{\sqrt{x^2 + 4x + 1}} + \frac{x+2}{\sqrt{x^2 + 4x + 1}} + C = \frac{x-2}{\sqrt{x^2 + 4x + 1}} + C$.

$\int \frac{2x + 1}{(x^2 + 4x + 1)^{3/2}} \, dx$

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