Question

Question #fcb11

Answer 1

You are correct: `intsqrt(x^2-1)/x^2dx=lnabs(x+sqrt(x^2-1))-sqrt(x^2-1)/x+C`

Explanation:

`I=intsqrt(x^2-1)/x^2dx`

I'm sure there are other methods of approach, but when I see square roots I immediately think of trig substitutions. Let's try `x=sectheta`, implying that `x^2-1=sec^2theta-1=tan^2theta` and that `dx=secthetatanthetad theta`:

`I=intsqrt(tan^2theta)/sec^2theta(secthetatanthetad theta)`

`I=inttan^2theta/secthetad theta`

`I=int(sec^2theta-1)/secthetad theta`

`I=int(sectheta-costheta)d theta`

`I=lnabs(sectheta+tantheta)-sintheta+C`

The substitution `x=sectheta` implies the following:

• `tantheta=sqrt(sec^2theta-1)=sqrt(x^2-1)`
• `costheta=1/x`
• `sintheta=sqrt(1-cos^2theta)=sqrt(1-1/x^2)=sqrt(x^2-1)/x`

Thus:

`I=lnabs(x+sqrt(x^2-1))-sqrt(x^2-1)/x+C`

You were right!