How do you find the integral #ln(x^2+4)#?

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Answer 1 · 2015-10-18T22:25:32

#I = xln(x^2+4)- 2x + 4arctan(x/2)+C#

#I=int ln(x^2+4)dx#

#u=ln(x^2+4) => du=1/(x^2+4)*2x*dx= (2x)/(x^2+4)dx#

#dv=dx => v=x#

#I=xln(x^2+4)- int x(2x)/(x^2+4)dx#

#I = xln(x^2+4)- 2int x^2/(x^2+4)dx#

#I = xln(x^2+4)- 2int (x^2+4-4)/(x^2+4)dx#

#I = xln(x^2+4)- 2int (1-4/(x^2+4))dx#

#I = xln(x^2+4)- 2(x-4*1/2arctan(x/2))+C#

#I = xln(x^2+4)- 2x + 4arctan(x/2)+C#