How do I evaluate $\int{\frac{x^3}{x^2-1}dx}$ using trigonometric substitution?

Question

How do I use trigonometric substitution to rewrite and evaluate

$$\int{\dfrac{x^3}{x^2-1}dx}?$$

I have no trouble using long division of $x^3$ by $x^2-1$ which is $x+\dfrac{x}{x^2-1}$ to get the answer $$\int \left(x+\dfrac{x}{x^2-1}\right)dx=\int x\ dx +\frac12\int \dfrac{2x}{x^2-1}dx$$$$=\frac{x^2}{2}+\frac12\ln|x^2-1|+c$$

but I'm stuck on figuring out an appropriate trigonometric substitution to rewrite this integral.

Any help would be appreciated!

Answer 1

Well, I guess you can put $\;x=\sin u\implies dx=\cos u\,du\;$ , so

$$\int\frac{x^3}{x^2-1}dx=\int\frac{\sin^3u}{\sin^2u-1}\cos u\,du=-\int\frac{\sin^3u}{\cos u}du=-\int\frac{\sin u(1-\cos^2)}{\cos u}du=$$

$$=-\int\frac{\sin u}{\cos u}du+\int \sin u\cos u\,du=\log|\cos u|=\frac12\sin^2u+C$$

You can now go back to $\;x\;$ if you want

Answer 2

Let $x=\sec\theta\implies dx=\sec\theta \tan\theta d\theta$ $$\int{\frac{x^3}{x^2-1}dx}=\int{\frac{\sec^3\theta}{\sec^2\theta-1}}\sec\theta \tan\theta d\theta$$ $$=\int \frac{\sec^4\theta d\theta}{\tan\theta}$$ $$=\int \frac{(1+\tan^2\theta)\sec^2\theta d\theta}{\tan\theta}$$

$$=\int \left(\frac{1}{\tan\theta}+\tan\theta\right)d(\tan \theta)$$ $$=\ln|\tan\theta|+\frac{\tan^2\theta}{2}+C$$ $$=\frac12\ln|x^2-1|+\frac{x^2}{2}+C$$

Answer 3

Of course you ask for a trigonometric answer. We can still see some old neat trick from "factoring"...:Observe that $\dfrac{x^3}{x^2-1} = \dfrac{x(x^2-1)+x}{x^2-1}= x+ \dfrac{x}{x^2-1}$,and the rest you should know what to do...

Answer 4

It dosen't make sense to use trig substitution: $$I=\int{\frac{x^3}{x^2-1}dx}$$ $$2I=\int{\frac{x^2}{x^2-1}dx^2}$$ Substitute $x=\sin u$: $$2I=\int{\frac{\sin^2 u}{\sin^2 u-1}d\sin ^2 u}$$ $$2I=\int{\frac{\sin^2 u-1+1}{\sin^2 u-1}d\sin ^2 u}$$ $$2I=\int d\sin^2 u+\int{\frac{1}{\sin^2 u-1}d\sin ^2 u}$$ $$2I=\sin^2 u+\ln |{\sin^2 u-1}|+C$$ $$ I=\dfrac 12 (x^2 +\ln |{ x^2-1}|)+C$$

Answer 5

Let $x=\cosh t$. Then,

$$\int{\frac{x^3}{x^2-1}dx}=\int \frac{\cosh^3t}{\sinh t} dt =\int (\sinh t + \frac1{\sinh t}) d(\sinh t)\\ = \frac12 \sinh^2 t + \ln|\sinh t|+C $$