How do you integrate $ln(x^2+4)$ ?

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Answer 1 · 2018-04-06T19:39:33

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Answer 2 · 2018-04-06T19:48:41

$xln(x^2+4)-2x+4arc tan(x/2)+C.$

We will use the following Rule of Integration by Parts :

$intu*vdx=uintvdx-int{(du)/dx*intvdx}dx$ .

We take, $u=ln(x^2+4) rArr (du)/dx=1/(x^2+4)*2x, and,$

$v=1 :. intvdx=x$ .

$:. int{ln(x^2+4)*1}dx=xln(x^2+4)-int{(2x)/(x^2+4)*x}dx$ ,

$=xln(x^2+4)-2intx^2/(x^2+4)dx$ ,

$=xln(x^2+4)-2int{(x^2+4)-4}/(x^2+4)dx$ ,

$=xln(x^2+4)-2int{(x^2+4)/(x^2+4)-4/(x^2+4)}dx$ ,

$=xln(x^2+4)-2{intdx-4int1/(x^2+2^2)dx}$ ,

$=xln(x^2+4)-2{x-4*1/2arc tan(x/2)}$ ,

$=xln(x^2+4)-2x+4arc tan(x/2)+C.$

How do you integrate ln(x^2+4)ln(x^2+4)?