int frac{x^2 operatorname{Tan}^{-1} x}{(1+x^2 | vedclass

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(A) ધારો કે $I = \int \frac{x^2 \operatorname{Tan}^{-1} x}{(1+x^2)^2} dx$. ખંડશઃ સંકલનનો ઉપયોગ કરતા, $u = \operatorname{Tan}^{-1} x$ અને $dv = \frac{x

Vedclass · Accepted Answer

$\frac{(\operatorname{Tan}^{-1} x)^2}{4} - \frac{x \operatorname{Tan}^{-1} x}{2(1+x^2)} + \frac{1-x^2}{4(1+x^2)} + c$

Vedclass · Answer

(A) ધારો કે $I = \int \frac{x^2 \operatorname{Tan}^{-1} x}{(1+x^2)^2} dx$. ખંડશઃ સંકલનનો ઉપયોગ કરતા, $u = \operatorname{Tan}^{-1} x$ અને $dv = \frac{x^2}{(1+x^2)^2} dx$ લો. તેથી $du = \frac{1}{1+x^2} dx$. $v$ શોધવા માટે, $\int \frac{x^2}{(1+x^2)^2} dx = \int \frac{x^2+1-1}{(1+x^2)^2} dx = \int \frac{1}{1+x^2} dx - \int \frac{1}{(1+x^2)^2} dx$. $x = 	an 	heta$ આદેશ લેતા, $dx = \sec^2 	heta d	heta$, આપણને મળે $\int \frac{1}{(1+x^2)^2} dx = \int \frac{\sec^2 	heta}{\sec^4 	heta} d	heta = \int \cos^2 	heta d	heta = \frac{1}{2} \int (1 + \cos 2	heta) d	heta = \frac{1}{2} (	heta + \frac{\sin 2	heta}{2}) = \frac{1}{2} \operatorname{Tan}^{-1} x + \frac{x}{2(1+x^2)}$. આમ, $v = \operatorname{Tan}^{-1} x - (\frac{1}{2} \operatorname{Tan}^{-1} x + \frac{x}{2(1+x^2)}) = \frac{1}{2} \operatorname{Tan}^{-1} x - \frac{x}{2(1+x^2)}$. હવે, $I = uv - \int v du = \operatorname{Tan}^{-1} x (\frac{1}{2} \operatorname{Tan}^{-1} x - \frac{x}{2(1+x^2)}) - \int (\frac{1}{2} \operatorname{Tan}^{-1} x - \frac{x}{2(1+x^2)}) \frac{1}{1+x^2} dx$. $I = \frac{(\operatorname{Tan}^{-1} x)^2}{2} - \frac{x \operatorname{Tan}^{-1} x}{2(1+x^2)} - \frac{1}{2} \int \frac{\operatorname{Tan}^{-1} x}{1+x^2} dx + \frac{1}{2} \int \frac{x}{(1+x^2)^2} dx$. $I = \frac{(\operatorname{Tan}^{-1} x)^2}{2} - \frac{x \operatorname{Tan}^{-1} x}{2(1+x^2)} - \frac{(\operatorname{Tan}^{-1} x)^2}{4} - \frac{1}{4(1+x^2)} + c$. $I = \frac{(\operatorname{Tan}^{-1} x)^2}{4} - \frac{x \operatorname{Tan}^{-1} x}{2(1+x^2)} - \frac{1}{4(1+x^2)} + c$.

$\int \frac{x^2 \operatorname{Tan}^{-1} x}{(1+x^2)^2} dx =$

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