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log-2-x-2-dx-




Question Number 27282 by hp killer last updated on 04/Jan/18
∫log(2+x^2 )dx
log(2+x2)dx
Commented by abdo imad last updated on 04/Jan/18
if you mean ln integrate par parts u^′ =1 and v=ln(2+x^2 )  ∫ln(2+x^2 )dx= xln(2 +x^2 ) −∫ x ((2x)/(2+x^2 ))dx  = xln(2+x^2 ) −2 ∫ (x^2 /(2+x^2 ))dx  =xln(2+x^2 ) −2 ∫((2+x^2  −2)/(2+x^2 ))dx  =xln(2+x^2 )−2x  +4 ∫(dx/(2+x^2 ))  and by the changement x=(√2)t  ∫  (dx/(2+x^2 )) = ∫ (((√2)dt)/(2+2t^2 ))= ((√2)/2) artan((x/( (√2)))) so  ∫ln(2+x^2 )dx= xln(2+x^2 ) −2x + 2(√2) arctan((x/( (√2)))).
ifyoumeanlnintegrateparpartsu=1andv=ln(2+x2)ln(2+x2)dx=xln(2+x2)x2x2+x2dx=xln(2+x2)2x22+x2dx=xln(2+x2)22+x222+x2dx=xln(2+x2)2x+4dx2+x2andbythechangementx=2tdx2+x2=2dt2+2t2=22artan(x2)soln(2+x2)dx=xln(2+x2)2x+22arctan(x2).

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