int frac{1}{(x^2+1)^2} dx = . . . . . . | vedclass

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(B) Let $I = \int \frac{dx}{(x^2+1)^2}$.Substitute $x = 	an 	heta$,then $dx = \sec^2 	heta d 	heta$.$I = \int \frac{\sec^2 	heta d 	heta}{(	an^

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let $I = \int \frac{dx}{(x^2+1)^2}$.Substitute $x = 	an 	heta$,then $dx = \sec^2 	heta d 	heta$.$I = \int \frac{\sec^2 	heta d 	heta}{(	an^2 	heta + 1)^2} = \int \frac{\sec^2 	heta d 	heta}{\sec^4 	heta} = \int \cos^2 	heta d 	heta$.Using the identity $\cos^2 	heta = \frac{1 + \cos 2 	heta}{2}$,we get:$I = \frac{1}{2} \int (1 + \cos 2 	heta) d 	heta = \frac{1}{2} (	heta + \frac{\sin 2 	heta}{2}) + c$.Since $\sin 2 	heta = \frac{2 	an 	heta}{1 + 	an^2 	heta}$ and $	heta = 	an^{-1} x$,we have:$I = \frac{1}{2} 	an^{-1} x + \frac{1}{4} \cdot \frac{2x}{1 + x^2} + c = \frac{1}{2} 	an^{-1} x + \frac{x}{2(1 + x^2)} + c$.

$\int \frac{1}{(x^2+1)^2} dx = . . . . . .$

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