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Question Number 110964 by mathdave last updated on 01/Sep/20

$$solve$$$${\int }_0^1\frac{x^2\ln x}{{(1+x^2)}^3}dx$$

Answered by mathdave last updated on 01/Sep/20

$$enjoymysolution$$$$letx=\tan y,dx=(1+{\tan }^{2}y)dy$$$$I={\int }_0^{\frac{\pi }{4}}\frac{\tan y\ln \left(\tan y\right)}{{(1+{\tan }^{2}y)}^3}\times (1+{\tan }^{2}y)dy$$$$I={\int }_0^{\frac{\pi }{4}}\frac{{\tan }^{2}y\ln \left(\tan y\right)}{{\sec }^{4}y}dy={\int }_0^{\frac{\pi }{4}}\frac{{\tan }^{2}y\ln \left(\tan y\right){\cos }^{4}y}{1}dy$$$$I={\int }_0^{\frac{\pi }{4}}{\left(\sin y\cos y\right)}^2\ln \left(\tan y\right)dy$$$$but\sin y\cos y=\frac{\sin 2y}{2}$$$$I=\frac{1}{4}{\int }_0^{\frac{\pi }{4}}{\sin }^{2}2y\ln \left(\tan y\right)dy$$$$but\cos 4y=1-{2\sin }^{2}2y,{\sin }^{2}2y=\frac{1-\cos 4y}{2}$$$$I=\frac{1}{8}{\int }_0^{\frac{\pi }{4}}(1-\cos 4y)\ln \left(\tan y\right)dy$$$$I=\frac{1}{8}{\int }_0^{\frac{\pi }{4}}\ln \left(\tan y\right)-\frac{1}{8}{\int }_0^{\frac{\pi }{4}}\cos 4y\ln \left(\tan y\right)dy$$$$letB={\int }_0^{\frac{\pi }{4}}\cos 4y\ln \left(\tan y\right)dy$$$$usingIBPbutlet$$$$\int dv=\int \cos 4ydy,v=\frac{\sin 4y}{4},u=\ln \left(\tan y\right),du=\frac{{\sec }^{2}y}{\tan y}=\frac{1}{\sin y\cos y}=\frac{2}{\sin 2y}$$$$B=\frac{1}{4}{\left[\sin 4y\ln \left(\tan y\right)\right]}_0^{\frac{\pi }{4}}-\frac{1}{4}{\int }_0^{\frac{\pi }{4}}\frac{2\sin 4y}{\sin 2y}dy$$$$B=-\frac{1}{2}{\int }_0^{\frac{\pi }{4}}\frac{\sin 4y}{\sin 2y}dy(let2y=x,dy=\frac{dx}{2})$$$$=-\frac{1}{4}{\int }_0^{\frac{\pi }{2}}\frac{\sin 2x}{\sin x}dx=-\frac{1}{4}{\int }_0^{\frac{\pi }{2}}\frac{2\sin x\cos x}{\sin x}dx$$$$B=-\frac{1}{2}{\int }_0^{\frac{\pi }{2}}\cos xdx=-\frac{1}{2}{\left[\sin x\right]}_0^{\frac{\pi }{2}}=-\frac{1}{2}$$$$\because B={\int }_0^{\frac{\pi }{4}}\cos 4y\ln \left(\tan y\right)dy=-\frac{1}{2}.........\left(1\right)$$$$thenletA={\int }_0^{\frac{\pi }{4}}\ln \left(\tan y\right)dy$$$$letx=\tan y,dy=\frac{dx}{1+x^2}$$$$A={\int }_0^1\frac{\ln x}{1+x^2}dx\left(usingseriessummation\right)$$$$A={\int }_0^1{(-1)}^n\underset{n=0}{\sum ^{\mathrm{\infty }}}{x}^{2n}\ln xdx={(-1)}^n\underset{n=0}{\sum ^{\mathrm{\infty }}}{\int }_0^1{x}^{2n}.x^{a}dx{\mid }_{a=0}$$$$\frac{\partial }{\partial a}{\mid }_{a=0}A\left(a\right)={(-1)}^n\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{\partial }{\partial a}{\int }_0^1{x}^{2n+a}dx$$$$\frac{\partial }{\partial a}{\mid }_{a=0}A\left(a\right)={(-1)}^n\frac{\partial }{\partial a}\underset{n=0}{\sum ^{\mathrm{\infty }}}\left[\frac{1}{(2n+a+1)}\right]$$$$A\left(0\right)=A=-\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{{(2n+1)}^2}=-G\left(catalanconstant\right).....\left(2\right)$$$$butI=\frac{1}{8}A-\frac{1}{8}B=[-\frac{G}{8}-\frac{1}{8}(-\frac{1}{2})]=[\frac{1}{16}-\frac{1}{8}G]$$$$\because {\int }_0^1\frac{x^2\ln x}{{(1+x^2)}^3}dx=[\frac{1}{16}-\frac{1}{8}G]$$$$solutionbymathdave\left(1/09/2020\right)$$

Commented by mnjuly1970 last updated on 01/Sep/20

$$okay..$$

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