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.#