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

2 Answers
Apr 14, 2018

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

Apr 14, 2018

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 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!