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Question Number 154994 by mathdanisur last updated on 24/Sep/21

$$\mathrm{Find}$$$$\underset{0}{\stackrel{1}{\int }}{\mathrm{x}}^2{\tan }^{-1}\left(2\mathrm{x}\right){\ln }^2\left(3\mathrm{x}\right)\mathrm{dx}=?$$

Answered by phanphuoc last updated on 24/Sep/21

$$I={ln}^{2}3{\int }_0^1{x}^{2}arctan\left(2x\right)dx+{\int }_0^1{x}^{2}arctan\left(2x\right)lnxdx+{\int }_0^1{x}^{2}arctan\left(2x\right){ln}^{2}xdx=$$$$=I_1+I_2+I_3$$$$J\left(a\right)={\int }_0^1{x}^{a}arctan\left(2x\right)dx$$$$u=arctan2x,dv=x^a\to du=2/(4x^2+1),v=x^{a+1}/(a+1)$$$$\to J=.......$$$$I_1=J\left(2\right),I_2=(J{\left(2\right)}^{\prime },I_3={\left(J\left(2\right)\right)}^{″}$$

Commented by mathdanisur last updated on 24/Sep/21

$$\mathrm{Thankyou}\mathbf{S}\mathrm{er},\mathrm{how}\mathrm{please}$$

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