how to solve $\int \frac{x^2}{\sqrt{x^2-13}}\text{d}x$?

Question

As the question says, I am trying to solve $\int \frac{x^2}{\sqrt{x^2-13}}\text{d}x$

Here is what I have so far:

$\int \frac{x^2}{\sqrt{x^2-13}}\text{d}x$

$\frac{1}{\sqrt{13}} \int \frac{x^2}{\sqrt{\frac{x^2}{13}-1}} \text{d}x$

$u = \frac{x}{\sqrt{13}}$

$\frac{\text{d}u}{\text{d}x} = \frac{1}{\sqrt{13}}$

$\frac{1}{\sqrt{13}}*\sqrt{13} \int \frac{13u^2}{\sqrt{u^2-1}}\text{d}u$

I got the $13u^2$ by setting $\sqrt{13}u = x$ and squaring both sides

$13 \int \frac{u^2}{\sqrt{u^2-1}}\text{d}u$

$u = \sec(\theta)$

$\frac{\text{d}u}{\text{d}\theta} = \sec(\theta)\tan(\theta)$

$13 \int \frac{\sec^2(\theta)}{\tan(\theta)}*\sec(\theta)\tan(\theta)\text{d}\theta$

$13 \int \sec^3(\theta)\text{d}\theta$

This is the point where I am stuck at

Also, I just started learning trig substitution so if there is an easier way to do this please tell me. Thank you

Answer 1

$$ x = \sqrt{13} \cosh t \; , \; \; \; dx = \sqrt{13} \sinh t dt $$ $$ \sqrt{x^2 - 13} = \sqrt{13 \cosh^2 t - 13 } = \sqrt {13} \sqrt{\cosh^2 t - 1} = \sqrt {13} \sinh t $$

$$ \int \frac{x^2}{\sqrt{x^2 - 13}} dx = 13 \int \cosh^2 t dt = 13 \int \frac{1}{2} + \frac{1}{2} \,\cosh 2t \, dt $$

$$ \int \frac{x^2}{\sqrt{x^2 - 13}} dx = C + 13 \left( \frac{t}{2} + \frac{1}{4} \,\sinh 2t\right) $$

now we need to express $t$ and $\sinh 2t $ in terms of $x$

I guess the quickest way comes from $ \cosh t + \sinh t = e^t$ So

$$ \sqrt{13 }e^t = x + \sqrt{x^2 - 13} $$

Also $$ x \, \sqrt{x^2 - 13} = 13 \sinh t \cosh t $$ $$ 2 x \, \sqrt{x^2 - 13} = 13 \sinh 2t $$

Answer 2

Hint

Let's try integration by parts: $$\int\sec^3\theta =\tan\theta \sec\theta-\int\sec\theta\tan^2\theta=\tan\theta\sec\theta-\int(\sec^3\theta-\sec\theta)\implies \int\sec^3\theta =\dfrac 12(\tan\theta \sec\theta +\int\sec\theta ).$$

Answer 3

$sec^3(\theta)=\frac{1}{cos^3(\theta)}=\frac{cos(\theta)}{cos^4(\theta)}=\frac{cos(\theta)}{(1-sin^2(\theta))^2}$.

Take $t=sin(\theta)$, then $$I=\int \frac{dt}{(1-t^2)^2}$$ $$\frac{1}{(1-t^2)^2}=\frac{1}{2}\frac{1-t+1+t}{(1-t)^2(1+t)^2}=\frac{1}{2}(\frac{1}{(1-t)(1+t)^2}+\frac{1}{(1-t)^2(1+t)})=\frac{1}{4}(\frac{1-t+1+t}{(1-t)(1+t)^2}+\frac{1-t+1+t}{(1-t)^2(1+t)})=\frac{1}{4}(\frac{1}{(1+t)^2}+\frac{2}{(1-t)(1+t)}+\frac{1}{(1-t)^2})=\frac{1}{4}(\frac{1}{(1+t)^2}+\frac{1-t+1+t}{(1-t)(1+t)}+\frac{1}{(1-t)^2})=\frac{1}{4}(\frac{1}{(1+t)^2}+\frac{1}{(1-t)}+\frac{1}{(1+t)}+\frac{1}{(1-t)^2}) $$ Hence, $$I=\frac{1}{4}(\frac{1}{1-t}-\frac{1}{1+t}+ln(1+t)-ln(t-1))$$

Answer 4

First of all, we deal with the integral by integration by parts. $$ \begin{aligned} I & =\int x^2 d\left(\sqrt{x^2-13}\right) \\ & =x \sqrt{x^2-13}-\int \sqrt{x^2-13}\ d x\\&= x \sqrt{x^2-13}-\int \frac{x^2-13}{\sqrt{x^2-13}} d x \end{aligned} $$ Rearranging yields

$$ I=\frac{1}{2} x \sqrt{x^2-13}+\frac{13}{2} \int \frac{d x}{\sqrt{x^2-13}} $$

Using the substitution $x=\sqrt 13 \sec \theta $, we get

$$ \begin{aligned} \int \frac{d x}{\sqrt{x^2-13}} & =\int \sec \theta d \theta \\ & =\ln |\sec \theta+\tan \theta|+c_1 \\ &= \ln \left|x+\sqrt{x^2-13}\right|+C \end{aligned} $$ Hence $$ I=\frac{1}{2}\left[x \sqrt{x^2-13}+13 \ln \left|x+\sqrt{x^2-13}\right|\right]+C $$