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

1 Answer
Oct 18, 2015

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

Explanation:

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