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

Find  ∫_( 0) ^( 1)  x^2  tan^(-1) (2x)ln^2 (3x)dx=?

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

Answered by phanphuoc last updated on 24/Sep/21

I=ln^2 3∫_0 ^1 x^2 arctan(2x)dx+∫_0 ^1 x^2 arctan(2x)lnxdx+∫_0 ^1 x^2 arctan(2x)ln^2 xdx=  =I_1 +I_2 +I_3   J(a)=∫_0 ^1 x^a arctan(2x)dx  u=arctan2x,dv=x^a →du=2/(4x^2 +1),v=x^(a+1) /(a+1)  →J=.......  I_1 =J(2),I_2 =(J(2)′,I_3 =(J(2))′′

$${I}={ln}^{\mathrm{2}} \mathrm{3}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{2}} {arctan}\left(\mathrm{2}{x}\right){dx}+\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{2}} {arctan}\left(\mathrm{2}{x}\right){lnxdx}+\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{2}} {arctan}\left(\mathrm{2}{x}\right){ln}^{\mathrm{2}} {xdx}= \\ $$$$={I}_{\mathrm{1}} +{I}_{\mathrm{2}} +{I}_{\mathrm{3}} \\ $$$${J}\left({a}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{a}} {arctan}\left(\mathrm{2}{x}\right){dx} \\ $$$${u}={arctan}\mathrm{2}{x},{dv}={x}^{{a}} \rightarrow{du}=\mathrm{2}/\left(\mathrm{4}{x}^{\mathrm{2}} +\mathrm{1}\right),{v}={x}^{{a}+\mathrm{1}} /\left({a}+\mathrm{1}\right) \\ $$$$\rightarrow{J}=....... \\ $$$${I}_{\mathrm{1}} ={J}\left(\mathrm{2}\right),{I}_{\mathrm{2}} =\left({J}\left(\mathrm{2}\right)',{I}_{\mathrm{3}} =\left({J}\left(\mathrm{2}\right)\right)''\right. \\ $$

Commented by mathdanisur last updated on 24/Sep/21

Thankyou Ser, how please

$$\mathrm{Thankyou}\:\boldsymbol{\mathrm{S}}\mathrm{er},\:\mathrm{how}\:\mathrm{please} \\ $$

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