intsqrt(1+1/x^2)dx=intsqrt((x^2+1)/x^2)dx=intsqrt(x^2+1)/xdx∫√1+1x2dx=∫√x2+1x2dx=∫√x2+1xdx
Letting u=sqrt(x^2+1)u=√x2+1 reveals that du=x/sqrt(x^2+1)dxdu=x√x2+1dx. Then the integral can be manipulated to become:
intsqrt(x^2+1)/xdx=int(x(x^2+1))/(x^2sqrt(x^2+1))dx=int(x^2+1)/x^2(x/sqrt(x^2+1)dx)∫√x2+1xdx=∫x(x2+1)x2√x2+1dx=∫x2+1x2(x√x2+1dx)
Note that u^2=x^2+1u2=x2+1 and u^2-1=x^2u2−1=x2:
int(x^2+1)/x^2(x/sqrt(x^2+1)dx)=intu^2/(u^2-1)du∫x2+1x2(x√x2+1dx)=∫u2u2−1du
Rewriting the integrand as (u^2-1+1)/(u^2-1)=1+1/(u^2-1)u2−1+1u2−1=1+1u2−1:
intu^2/(u^2-1)du=intdu+int1/(u^2-1)du=u+int1/(u^2-1)du∫u2u2−1du=∫du+∫1u2−1du=u+∫1u2−1du
You can perform partial fraction decomposition on 1/(u^2-1)1u2−1 to see that 1/(u^2-1)=1/(2(u-1))-1/(2(u+1))1u2−1=12(u−1)−12(u+1):
u+int1/(u^2-1)du=u+1/2int1/(u-1)du-1/2int1/(u+1)duu+∫1u2−1du=u+12∫1u−1du−12∫1u+1du
Both of which you could perform a substitution on, or just recognize them for natural log integrals:
u+1/2int1/(u-1)du-1/2int1/(u+1)du=u+1/2lnabs(u-1)-1/2lnabs(u+1)u+12∫1u−1du−12∫1u+1du=u+12ln|u−1|−12ln|u+1|
Combining:
=u+1/2lnabs(u-1)-1/2lnabs(u+1)=u+1/2lnabs((u-1)/(u+1))=u+12ln|u−1|−12ln|u+1|=u+12ln∣∣∣u−1u+1∣∣∣
Since u=sqrt(x^2+1)u=√x2+1:
u+1/2lnabs((u-1)/(u+1))=sqrt(x^2+1)+1/2lnabs((sqrt(x^2+1)-1)/(sqrt(x^2+1)+1))u+12ln∣∣∣u−1u+1∣∣∣=√x2+1+12ln∣∣∣√x2+1−1√x2+1+1∣∣∣
Or:
intsqrt(1+1/x^2)dx=sqrt(x^2+1)+1/2lnabs((sqrt(x^2+1)-1)/(sqrt(x^2+1)+1))+C∫√1+1x2dx=√x2+1+12ln∣∣∣√x2+1−1√x2+1+1∣∣∣+C
