Problem 7.5.29

I wasn't able to solve this problem in the first section, but we were on the right track! A student in my second section came up with the solution.

Problem:
$$\int \ln(x+\sqrt{x^2-1})dx$$

Because of the term $\sqrt{x^2-1}$ our intuition told us to try a trig substitution. We let $x=\sec\theta$, and obtained $$ \int \ln \left( \sec\theta+\tan\theta \right) \sec\theta \tan\theta d\theta$$. What I failed to recognize, was that $\ln \left|\sec\theta +\tan\theta \right|$ is the integral of $\sec \theta$ with respect to $\theta$ (page 475).

In any case, now we can integrate by parts. $u= \ln \left( \sec \theta +\tan\theta \right)$ and $dv=\sec\theta\tan\theta d\theta$. So that $du= \sec\theta d\theta$ and $v=\sec\theta$. Applying integration by parts gives $$\sec \theta \ln \left(\sec\theta+\tan\theta\right)-\int \sec^2 \theta d\theta$$

And since $\int \sec ^2 \theta d\theta = \tan \theta +C$, we get $$\sec \theta \ln\left(\sec\theta+\tan\theta\right)-\tan \theta + C$$

We need to return to the variable $x$. $\sec\theta=x$ and $\tan\theta =\sqrt{x^2-1}$. Thus $$x \ln \left(x+\sqrt{x^2-1}\right)-\sqrt{x^2-1}+C$$

We can check our answer by differentiating with respect to $x$.

$$\ln \left(x+\sqrt{x^2-1}\right)+\frac{x}{x+\sqrt{x^2-1}}\left(1+\frac{x}{\sqrt{x^2-1}}\right)-\frac{x}{\sqrt{x^2-1}}$$

This equals

$$\ln\left(x+\sqrt{x^2-1}\right)+\frac{x}{x+\sqrt{x^2-1}}+\frac{x}{\sqrt{x^2-1}}\frac{-\sqrt{x^2-1}}{x+\sqrt{x^2-1}}$$

$$\ln\left(x+\sqrt{x^2-1}\right)+\frac{x}{x+\sqrt{x^2-1}}+\frac{-x}{x+\sqrt{x^2-1}}$$

$$\ln\left(x+\sqrt{x^2-1}\right)$$