int frac{x}{(x^2+2x+2)^2} dx is equal to | vedclass

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

Vedclass · Accepted Answer

$-\frac{x+2}{2(x^2+2x+2)} - \frac{1}{2} 	an^{-1}(x+1) + C$

Vedclass · Answer

(A) Let $I = \int \frac{x}{(x^2+2x+2)^2} dx$. Rewrite the numerator as $x = \frac{1}{2}(2x+2) - 1$. Then $I = \frac{1}{2} \int \frac{2x+2}{(x^2+2x+2)^2} dx - \int \frac{1}{(x^2+2x+2)^2} dx$. For the first integral,let $u = x^2+2x+2$,so $du = (2x+2)dx$. $\frac{1}{2} \int u^{-2} du = -\frac{1}{2u} = -\frac{1}{2(x^2+2x+2)}$. For the second integral,let $J = \int \frac{1}{((x+1)^2+1)^2} dx$. Put $x+1 = 	an 	heta$,so $dx = \sec^2 	heta d	heta$. $J = \int \frac{\sec^2 	heta}{\sec^4 	heta} d	heta = \int \cos^2 	heta d	heta = \int \frac{1+\cos 2	heta}{2} d	heta = \frac{1}{2}(	heta + \frac{\sin 2	heta}{2}) + C$. Since $	an 	heta = x+1$,$	heta = 	an^{-1}(x+1)$ and $\sin 2	heta = \frac{2	an 	heta}{1+	an^2 	heta} = \frac{2(x+1)}{x^2+2x+2}$. So $J = \frac{1}{2} 	an^{-1}(x+1) + \frac{x+1}{2(x^2+2x+2)} + C$. Combining these,$I = -\frac{1}{2(x^2+2x+2)} - \frac{x+1}{2(x^2+2x+2)} - \frac{1}{2} 	an^{-1}(x+1) + C = -\frac{x+2}{2(x^2+2x+2)} - \frac{1}{2} 	an^{-1}(x+1) + C$.

$\int \frac{x}{(x^2+2x+2)^2} dx$ is equal to

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