int ln(x^2+4)dx
Use Integration by Parts
int u dv/dx = uv - int v du/dx
int ln(x^2+4)dx = int ln(x^2+4).1dx
u=ln(x^2+4)
du=2x/(x^2+4)dx using chain rule
dv=1
v=int(1)dx=x+constant
using the integration by parts formula
int ln(x^2+4)dx=xln(x^2+4) - intx(2x/(x^2+4))dx
=xln(x^2+4)- int(2x^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)- 2int(1-4/(x^2+4))dx
Now the standard integral int dx/(a^2+x^2) = 1/a tan^-1(x/a)
Therefore
=xln(x^2+4)- 2x+2(4int1/(x^2+2^2))dx ----equation1
=xln(x^2+4)- 2x+2(4/2tan^-1(x/2))
=xln(x^2+4)- 2x+4tan^-1(x/2)
If you can't remember the standard integral above then we can integrate this using a substitution
int4/(x^2+4)
x=2tanA
dx=2sec^2AdA
=int(2sec^2A)/(tan^2A+1)dA
sec^2A-1 = tan^2A
=int(2sec^2A)/(sec^2A)dA
=2A+constant
= 2tan^-1(x/2)
This can be substituted into equation 1
