Solve $\int \frac{x\ln(x)}{\sqrt{x^2 - 1}}$

Question

My Work

$x = \tan\theta$

$dx = \sec^2\theta d\theta$

$\int \frac{\tan\theta\ln(\tan\theta)}{\sqrt{\tan^2\theta - 1}}\sec^2\theta d\theta$

$\int \sec\theta \tan\theta ln(tan\theta)$

$u = \ln(\tan\theta)$

$du = \frac{\sec^2 \theta}{\tan \theta}$

$dv = \sec \theta \tan \theta d\theta$

$v = \sec \theta$

$\sec\theta\ln\tan\theta - \int \frac{\sec^3\theta}{\tan \theta}$

$\sec\theta\ln\tan\theta - \int \sec^2\theta \csc\theta d\theta$

I'm stuck after here. Parts doesn't look particularly appealing. I don't see an easy substitution. Brain is pretty tired at this point. Anyone know what to do?

Answer 1

A start: Integrate by parts, letting $u=\ln x$ and $dv=\frac{x}{\sqrt{x^2-1}}\,dx$.

Remark: The substitution $x=\tan\theta$ is not good for $\sqrt{x^2-1}$. There the appropriate thing is $x=\sec\theta$ or $x=\cosh t$.

Answer 2

First integrate by parts, then substitute $x=\sec{\theta}$ $\implies \mathrm{dx}=\tan{\theta}\sec{\theta}\,\mathrm{d}\theta$:

$$\begin{align} \int\frac{x\,\ln{\left(x\right)}}{\sqrt{x^2-1}}\,\mathrm{d}x &=\sqrt{x^2-1}\,\ln{\left(x\right)}-\int\frac{\sqrt{x^2-1}}{x}\,\mathrm{d}x\\ &=\sqrt{x^2-1}\,\ln{\left(x\right)}-\int\frac{\sqrt{\sec^2{\theta}-1}}{\sec{\theta}}\,\tan{\theta}\sec{\theta}\,\mathrm{d}\theta\\ &=\sqrt{x^2-1}\,\ln{\left(x\right)}-\int\tan^2{\theta}\,\mathrm{d}\theta\\ &=\sqrt{x^2-1}\,\ln{\left(x\right)}-\int\left(\sec^2{\theta}-1\right)\,\mathrm{d}\theta\\ &=\sqrt{x^2-1}\,\ln{\left(x\right)}-\tan{\theta}+\theta+\color{grey}{constant}\\ &=\sqrt{x^2-1}\,\ln{\left(x\right)}-\sqrt{x^2-1}+\sec^{-1}{x}+\color{grey}{constant}\\ \end{align}$$

Answer 3

You can evaluate the integral without using secant substitution. Here is the way.

Since we know that \begin{gather*} \mathrm{d}\sqrt{x^2-1}=\frac{x}{\sqrt{x^2-1}}\mathrm{d} x, \end{gather*} we can calculate, by differentiation by parts, as follows, \begin{align*} &\quad \frac{x\ln(x)}{\sqrt{x^2-1}} \mathrm{d} x=\ln(x)\mathrm{d} \sqrt{x^2-1}=\mathrm{d} \big(\ln(x)\sqrt{x^2-1}\big)-\frac{\sqrt{x^2-1}}{x}\mathrm{d} x\\ &= \mathrm{d} \big(\ln(x)\sqrt{x^2-1}\big)-\frac{x^2-1}{x\sqrt{x^2-1}}\mathrm{d} x\\ &=\mathrm{d} \big(\ln(x)\sqrt{x^2-1}\big)-\frac{x}{\sqrt{x^2-1}}\mathrm{d} x+\frac{1}{x\sqrt{x^2-1}}\mathrm{d} x\\ &=\mathrm{d} \big(\ln(x)\sqrt{x^2-1}\big)-\mathrm{d}\big(\sqrt{x^2-1}\big)+\frac{1}{x^2\sqrt{1-\left(\frac{1}{x}\right)^2}}\mathrm{d} x\\ &=\mathrm{d} \big(\ln(x)\sqrt{x^2-1}\big)-\mathrm{d}\big(\sqrt{x^2-1}\big)-\frac{1}{\sqrt{1-\left(\frac{1}{x}\right)^ 2}}\mathrm{d} \left(\frac{1}{x}\right)\\ &=\mathrm{d}\left(\ln(x)\sqrt{x^2-1}-\sqrt{x^2-1}-\arcsin\left(\frac{1}{x}\right)\right), \end{align*} hence we have \begin{gather*} \int\frac{x\ln(x)}{\sqrt{x^2-1}}\mathrm{d} x=\ln(x)\sqrt{x^2-1}-\sqrt{x^2-1}-\arcsin\left(\frac{1}{x}\right)+C. \end{gather*}