Any simple integration to this indefinite integral?

Question

$I =\displaystyle\int \dfrac{\sqrt{4+9x^4}}{x^3}dx$

One method we have tried is to use the substitution $x^2=\displaystyle\frac2{3\tan\theta}$ ,but it seems hard to change back the $\theta$ to x in the answer.

Update:

My answer using the substituion $x^2=\displaystyle\frac2{3\tan\theta}$

$I =\displaystyle\int \dfrac{\sqrt{4+9x^4}}{x^3}dx$
$=\displaystyle\int {\sqrt{4+9(\frac{4}{9\tan^2\theta})}(\sec^2\theta)(-\frac{3}{4}) d\theta}$
$=\displaystyle-\frac{3}{2}\int {\sqrt{\frac{1+\tan^2\theta}{\tan^2\theta}}\sec^2\theta d\theta}$
$=\displaystyle-\frac{3}{2}\int {\frac{\sec^3\theta}{\tan\theta}d\theta}$
$=\displaystyle-\frac{3}{2}\int {\frac{\sin\theta}{\cos^2\theta\sin^2\theta}d\theta}$
$=\displaystyle-\frac{3}{2}\int {\frac{\sin\theta}{\cos^2\theta(1-\cos^2\theta)}d\theta}$
$=\displaystyle-\frac{3}{2}\int {\frac{1}{(1-t^2)t^2}(-dt)}$ by letting $t=\cos\theta$
$=\displaystyle\frac{3}{2}\int \big({\frac{1}{1-t^2}+\frac{1}{t^2}\big)dt}$
$=\displaystyle\frac{3}{2}\int \big[\big(\frac{1}{2})({\frac{1}{1+t}+\frac{1}{1-t})+\frac{1}{t^2}\big]dt}$
$=\displaystyle\frac{3}{4} \big(\ln({1+t})-\ln({1-t})-\frac{2}{t}\big)$
$=\displaystyle\frac{3}{4} \big(\ln({1+\cos\theta})-\ln({1-\cos\theta})-2\sec\theta\big)$

and then substitute back $\displaystyle\cos\theta = \frac{3x^2}{\sqrt{4+9x^2}}$ into the answer

Answer 1

Let $x^2=u\implies dx=?$

$$I=2\int\dfrac{\sqrt{4+9u^2}}{u^2}du$$

Now integrate by parts $$I=2\sqrt{4+9u^2}\int\dfrac{du}{u^2}-2\int\left(\dfrac{d\sqrt{4+9u^2}}{du}\int\dfrac{du}{u^2}\right)du=?$$

Answer 2

Hint...you could try substituting $$x^2=\frac 23 \sinh u$$

$$\implies 2xdx=\cosh u du$$

and $$4+9x^4=4+4\sinh^2u=4\cosh^2u$$

Answer 3

Integrate by parts \begin{align}\int \frac{\sqrt{4+9x^4}}{x^3}dx &= \int {\sqrt{4+9x^4}}\ d\left(-\frac1{2x^2}\right)\\ &=-\frac{\sqrt{4+9x^4}}{2x^2}+\frac94\int \frac{d(x^2)}{\sqrt{1+\frac94 x^4}}\\ &=-\frac{\sqrt{4+9x^4}}{2x^2}+\frac32\sinh^{-1}\frac{3x^2}2+C \end{align}