Integrate $\int \frac {x^4}{\sqrt {x^2-9}} \,dx$

Question

Integrate $\int \dfrac {x^4}{\sqrt {x^2-9}} dx$

My Attempt: Let $x=3\sec (\theta )$ $$dx=3\sec (\theta).\tan (\theta).d\theta$$ Then, $$=\int \dfrac {x^4}{\sqrt {x^2-9}}$$ $$=\int \dfrac {81. \sec^4 (\theta)}{\sqrt {(3\sec (\theta))^2 - 9}} 3\sec (\theta).\tan (\theta).d\theta $$ $$=\int \dfrac {81 \sec^5 (\theta). 3\tan (\theta).d\theta}{3\tan (\theta)}$$ $$=81\int \sec^5 (\theta) d\theta$$

Answer 1

To integrate $\int \sec^5 (\theta) d\theta$, use integration by parts.

u = $\sec^3 (\theta)$

dv = $\sec^2 (\theta) d\theta$

And so du = $\sec^3\theta\tan\theta d \theta$

and $v = \tan\theta$

Thus $$\int \sec^5 (\theta) d\theta = \sec^3 (\theta) \tan \theta - \int 3\sec^3\theta(sec^2\theta-1)d\theta = \sec^3 (\theta) \tan \theta - 3\int \sec^5\theta d\theta + 3\int \sec^3\theta d\theta$$

Move the $3\int \sec^5\theta d\theta$ to the left

$$4\int sec^5\theta d\theta = \sec^3\theta \tan \theta + 3\int sec^3 \theta d\theta$$

Now we have to find $\int \sec^3 \theta d\theta $, which can be done by using integration by parts as well. I skipped a step where you convert $\tan^2 \theta$ to $\sec^2 \theta -1$ so that one of the terms have an integral with secant to the power of the original integral you are looking for.

$$u = \sec \theta$$ $$dv = \sec^2 \theta d \theta$$

$$\int sec^3 \theta d \theta = \sec \theta \tan \theta - \int \sec \theta \tan^2 \theta d \theta$$

$$\int \sec^3 \theta d \theta = \sec \theta \tan \theta - \int \sec^3 \theta d \theta + \int \sec \theta d \theta $$

$$ 2\int \sec^3 \theta d\theta = \sec \theta \tan \theta + \int \sec \theta d \theta$$

$$ \int \sec^3 \theta d \theta = \frac{1}{2} \sec \theta \tan \theta + \frac{1}{2} \ln\left|sec \theta + \tan \theta \right| + C$$

Sub the value of $\int \sec^3 \theta d \theta $ back to the original equation, we get: $$ \int sec^5\theta d\theta = \frac{1}{4} \sec^3 \theta \tan \theta + \frac{3}{8} \sec \theta \tan \theta + \frac{3}{8} ln\lvert \sec \theta + \tan \theta \rvert + C $$

Since $\theta = arcsec \frac{x}{3}$, we can simply plug it in. Note that according to the Pythagorean theorem, $\cos \theta = \frac{3}{x}$ implies that $\tan \theta = \frac{\sqrt{x^2-9}}{3}$.

$$ \int sec^5\theta d\theta = \frac{x^3\sqrt{x^2-9}}{4*81} + \frac{x\sqrt{x^2-9}}{8*3} + \frac{3}{8}\ln \left| \frac{x+\sqrt{x^2-9}}{3}\right| + C$$

Multiply both sides by 81

$$ 81 \int sec^5\theta d\theta = \frac{x^3\sqrt{x^2-9}}{4} + \frac{27x\sqrt{x^2-9}}{8} + \frac{243}{8}\ln \left|\frac{x+\sqrt{x^2-9}}{3} \right|t + C$$

So the answer is:

$$ \int \dfrac {x^4}{\sqrt {x^2-9}} dx = \frac{x^3\sqrt{x^2-9}}{4} + \frac{27x\sqrt{x^2-9}}{8} + \frac{243}{8}\ln \left| \frac{x+\sqrt{x^2-9}}{3}\right| + C$$

Notice that you can split $\ln \left| \frac{x+\sqrt{x^2-9}}{3} \right|$ into the difference of two natural logs, but it is probably not necessary.

I found out that some people already posted the generalized form of this method, but this is more elaborate if you want the detailed steps.

Answer 2

Integrate by parts to establish the reduction formula $$\int \frac{x^n}{\sqrt{x^2-9}}dx =I_n= \frac{x^{n-1}}n\sqrt{x^2-9}+\frac{9(n-1)}nI_{n-2} $$ Apply the formula twice to reduce the integral $$\int \frac{x^4}{\sqrt{x^2-9}}dx=\left(\frac{x^3}4+\frac{27x}8\right) \sqrt{x^2-9} +\frac{243}8I_0 $$ where $ I_0=\int \frac{1}{\sqrt{x^2-9}}dx=\tanh^{-1}\frac{\sqrt{x^2-9}}x $.

Answer 3

Let $ x + t = \sqrt{x^2 - 9}$

$$I = \dfrac{-1}{16}\int \dfrac{(t^2 + 9)^4}{t^5} dt = -\dfrac{486\ln(|t|)+\dfrac{t^4+72t^2}{4}-\dfrac{5832t^2+6561}{4t^4}}{16}+C$$

Then, $$\dfrac1{16}\left(\dfrac{t^4+72t^2}{4}-\dfrac{5832t^2+6561}{4t^4}\right) =- \left(\dfrac{27}8 x \sqrt{x^2 - 9} + x^3\dfrac{1}4 \sqrt{x^2 - 9}\right) $$

$$I = \dfrac{27}8 x \sqrt{x^2 - 9} + x^3\dfrac{1}4 \sqrt{x^2 - 9} - \dfrac{243}8\ln(\sqrt{x^2 - 9} - x) + C$$

Answer 4

For $x\ge 3$ let substitute $\begin{cases}x=3\cosh(u) & u\ge 0\\3\sinh(u)=\sqrt{x^2-9}\end{cases}$

$\begin{align}\displaystyle \int \dfrac{x^4}{\sqrt{x^2-9}}\mathop{dx}&=81\int \cosh(u)^4\mathop{du}\\&=\dfrac{81}8\bigg(2\cosh(u)^3\sinh(u)+3\cosh(u)\sinh(u)+3u\bigg)+C\\&=\frac 14x^3\sqrt{x^2-9}+\dfrac{27}{8}x\sqrt{x^2-9}+\dfrac{243}8\operatorname{argch}(\frac x3)+C\end{align}$

For $x\le-3$ just reverse the sign.

Your choice of substitution leads to $\sec^5(x)$ which is difficult to linearise.

With $\cosh^4(u)=\frac 18\cosh(4u)+\frac 12\cosh(2u)+\frac 38$ we can integrate and apply the inverse transformation to go back to powers of $(\cosh,\sinh)$, I wrote only the resulting anti-derivative.