Evaluating an indefinite integral with an inverse trigonometric function

I'm really stumped on a homework problem asking me to evaluate $\int \frac{ln\ 6x\ sin^{-1}(ln6x)}{x}dx$, and after a few hours of trying different approaches I'd definitely be appreciative for a bump in the right direction. As a caveat, I should add that I've already received credit for the assignment, I'm simply looking to fully understand how to complete the integral.

Here's what I've done so far:

I noticed a good u-substitution, so I let $u = ln\ 6x$ and $du = \frac{1}{x}dx$

So I rewrote my integral as $\int u\ sin^{-1}(u)\ du$

This particular section is allowing me to use formulas for integration so I've chosen:

$$\int x^n sin^{-1}x\ dx = \frac{1}{n+1} \left(x^{n+1}sin^{-1}x-\int\frac{x^{n+1}dx}{\sqrt{1-x^2}}\right)$$

Which gets me:

$$\frac{1}{2} \left(u^2sin^{-1}u-\int\frac{u^2du}{\sqrt{1-u^2}} \right)$$

Now I use a second formula for integration which states:

$$\int \frac{x^2}{\sqrt{a^2-x^2}}dx = -\frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}sin^{-1}\frac{x}{a}+C$$

So this brings me to: \begin{align} &\frac{1}{2} \left(u^2sin^{-1}u-\left(-\frac{u}{2}\sqrt{1-u^2}+\frac{1}{2}sin^{-1}u\right)\right)+C \\ & = \frac{1}{4} \left( u \, \sqrt{1- u^{2}} + (2 u^{2} -1) \, \sin^{-1}(u) \right) + C \end{align} However, even when I replace $u$ with $ln\ 6x$ I can't seem to find a way to get to the answer, which is:

$$\frac{1}{4}\left((2 \, \ln^2(6x)-1) \, \sin^{-1}(\ln(6x)) + \ln(6x) \, \sqrt{1-\ln^2(6x)}\right)+C$$

Is my fundamental approach flawed or am I simply missing something in the latter stages of simplification?

• When you get to $\int\frac{u^2}{\sqrt{1-u^2}}du$, try making a trig substitution. – user84413 Oct 9 '14 at 17:06
• There were two minor errors, but it was easier to correct your result than type an almost identical result. In general, the process is correct. – Leucippus Oct 9 '14 at 17:51
• Thank you both, this helped a lot. – user153085 Oct 9 '14 at 20:03

Hints; for the second integral try the substitution $$u=\sin(t)$$
When you get $$\sin^2(t)$$ in the integrand, use the identity $$\sin^2(t)=\frac12 -\frac12 \cos(2t)$$