# Integration problem

I have been working on this problem:
$$\int xln{x}/\sqrt{x^2-1}}$$
but I haven't been able to come up with a solution.

First, I tried to solve it using integration by parts:
u= $$ln{x}$$ dv= $$x\sqrt{x^2-1}$$
du= $$\fracc{1/x}$$ v= $$\fracc{1/2} ln{x^2-1}$$

And arrived at:
$$\int {x\ln{x}\sqrt{x^2-1} = \fracc{1/2} \ln{x^2-1} - \fracc{1/2}\int \ln{x^2-1}/x$$
and that is where I got stuck.

So I chose to take another path and started over by letting:
x= $$\sec\Theta$$ dx= $$\sec\Theta\tan\Theta$$ dΘ

I substituted those values in the original integral and after simplifying I came up with:

$$\int {\sec^2\Theta} \ln\sec\Theta dΘ$$ .

And From there I tried to make a substitution for secΘ, but I was still not able to solve it. (having problems with the natural log expression).

Any help that would guide me to right the path to help me solve this problem would be greatly appreciated. Thanx!

Last edited:

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James R
Homework Helper
Gold Member
For a start, you don't have the right expression for v in your integration by parts.

$$\int \frac{x}{\sqrt{x^2 - 1}} dx = \sqrt{x^2 - 1}$$

You made an error in your integration by parts. $$\int \frac{x ln(x)}{\sqrt{x^2-1}} dx$$

u = ln(x) du = 1/x dx

dv = x/sqrt(x^2-1) dx v = below

$$\int \frac{x}{\sqrt{x^2-1}} dx$$

u = x^2 -1, du = 2x dx

$$\frac{1}{2} \int \frac{2x}{\sqrt{x^2-1}} dx =\frac{1}{2} \int \frac{1}{\sqrt{u}} du = \sqrt{u} = \sqrt{x^2-1}$$

So [itex] v = \sqrt{x^2-1} [/tex]

Note the u substitution in the second half is independent of the integration by parts. Try again form here.

Last edited:
Thanks guys! I was finally able to work it out with your help =)

The final answer I got was:
$$\ln{x}\sqrt{x^2-1} - \sqrt{x^2-1} + 2\arctan{\sqrt{x^2+1}}$$