Question

What is `intln(x^2-1)dx`?

Answer 1

`x ln(x^2-1)-2x+ln |(x+1)/(x-1)|+C`

Explanation:

We integrate this by parts

`int ln(x^2-1) dx = int ln(x^2-1)*1 dx`
`qquad = ln(x^2-1) int 1dx-int(d/dx(ln(x^2-1))times int 1dx)dx`
`qquad = x ln(x^2-1) -int ((2x)/(x^2-1)*x)dx`
`qquad = x ln(x^2-1)-int (2x^2-2+2)/(x^2-1)dx`
`qquad = x ln(x^2-1)-int 2dx+ int (1/(x+1)-1/(x-1))dx`
`qquad = x ln(x^2-1)-2x+ln |(x+1)/(x-1)|+C`

Answer 2

The final integral is `xln|x^2-1|+ln|(x+1)/(x-1)|-2x+C`.

Explanation:

Use the log properties to split up the integral:

`color(white)=intln(x^2-1)dx`

`=intln((x+1)(x-1))dx`

`=int(ln(x+1)+ln(x-1))dx`

`=intln(x+1)dx+intln(x-1)dx`

Say `I_1` is the left integral and `I_2` is the right one. Solve each one separately:

`I_1=intln(x+1)dx`

Let `u=x+1` which means `du=dx`:

`color(white)(I_1)=intlnu` `du`

Use the DI method (an easier way to understand integration by parts) to solve the integral:

`I_1=lnu*u-int 1/u*u` `du`

`color(white)(I_1)=u lnu-int 1du`

`color(white)(I_1)=u lnu-u+C`

`color(white)(I_1)=(x+1)ln(x+1)-(x+1)+C`

`color(white)(I_1)=(x+1)ln(x+1)-x-1+C`

`color(white)(I_1)=(x+1)ln(x+1)-x+C`

Do the same for the other integral:

`I_2=intln(x-1)dx`

Let `u=x-1` which means `du=dx`:

`color(white)(I_2)=intlnu` `du`

Same integral as before:

`color(white)(I_2)=u lnu-u+C`

`color(white)(I_2)=(x-1)ln(x-1)-(x-1)+C`

`color(white)(I_2)=(x-1)ln(x-1)-x+1+C`

`color(white)(I_2)=(x-1)ln(x-1)-x+C`

Combine the two integrals:

`color(white)=intln(x^2-1)dx`

`=I_1+I_2`

`=(x+1)ln(x+1)-x+C+(x-1)ln(x-1)-x+C`

`=(x+1)ln(x+1)+(x-1)ln(x-1)-2x+C`

You could leave this as your answer, or expand the multiplication and use log rules to simplify a little further:

`=xln(x+1)+ln(x+1)+xln(x-1)-ln(x-1)-2x+C`

`=x(ln(x+1)+ln(x-1))+ln(x+1)-ln(x-1)-2x+C`

`=xln(x^2-1)+ln((x+1)/(x-1))-2x+C`

Lastly, add absolute value bars to the `ln`s to increase the domain:

`=xln|x^2-1|+ln|(x+1)/(x-1)|-2x+C`

If you bring the `x` coefficient into the `ln` as an exponent, you can condense the integral a lot, looking like this:

`=ln(|x+1|^(x+1)|x-1|^(x-1))-2x+C`

That's the integral. Hope this helped!