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 lns 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!