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

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(B) Let $I = \int \frac{x 	an^{-1} x}{(1 + x^2)^{3/2}} \, dx$. Substitute $x = 	an 	heta$,then $dx = \sec^2 	heta \, d	heta$. Also,$1 + x^2 = 1 +

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let $I = \int \frac{x 	an^{-1} x}{(1 + x^2)^{3/2}} \, dx$. Substitute $x = 	an 	heta$,then $dx = \sec^2 	heta \, d	heta$. Also,$1 + x^2 = 1 + 	an^2 	heta = \sec^2 	heta$. Substituting these into the integral: $I = \int \frac{	an 	heta \cdot 	heta \cdot \sec^2 	heta}{(\sec^2 	heta)^{3/2}} \, d	heta = \int \frac{	heta 	an 	heta \sec^2 	heta}{\sec^3 	heta} \, d	heta = \int 	heta \frac{	an 	heta}{\sec 	heta} \, d	heta = \int 	heta \sin 	heta \, d	heta$. Using integration by parts: $I = 	heta(-\cos 	heta) - \int 1 \cdot (-\cos 	heta) \, d	heta = -	heta \cos 	heta + \sin 	heta + c$. Since $x = 	an 	heta$,we have $\cos 	heta = \frac{1}{\sqrt{1 + x^2}}$ and $\sin 	heta = \frac{x}{\sqrt{1 + x^2}}$. Substituting back: $I = -(	an^{-1} x) \frac{1}{\sqrt{1 + x^2}} + \frac{x}{\sqrt{1 + x^2}} + c = \frac{x - 	an^{-1} x}{\sqrt{1 + x^2}} + c$.

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

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