# Trigonometric integration question (tricky substitution)

Integral and result

\begin{eqnarray*} \int \frac{3x^2}{x^3 \sqrt{16x^6-17}} dx = \frac{1}{\sqrt{17}} \tan^{-1} \left(\frac{\sqrt{16x^6-17}}{\sqrt{17}} \right) +C \end{eqnarray*}

Is there a concise substitution to solve that integral? The integral calculator that I've used has multiple redundant steps that I found out about in other problems. I tried substituting $u=x^3$ but I am not sure how to handle the -17 in the root and convert the denominator into a positive expression to use the arctan integration.

Edit: Thanks for the answers. Substituting u as the square root helped me form he body of the answer. Also, the image I posted has an error in the answer; it is the sqrt(17) is only 17 in the denominator of the answer.

• have you tried $u=\left( 16x^6 - 17\right)$ – phdmba7of12 Jan 20 '18 at 23:47
• Kind of yeah. The integral calculator showed something like that but with the square root of u that you typed above (just like the answers below). – Nasser Jan 21 '18 at 10:17

I'd set $\sqrt{16x^6-17}=u$, so $16x^6-17=u^2$ and $$x^5\,dx=\frac{1}{48}u\,du \qquad x^6=\frac{u^2+17}{16}$$ Your integral is therefore $$\int\frac{3x^5}{x^6\sqrt{16x^6-17}}\,dx= \int\frac{1}{u^2+17}\,du$$
$$\int\frac{3x^2}{x^3\sqrt{16x^6-17}}\,dx=\int\frac{du}{u\sqrt{16u^2-17}}$$
$u=\sqrt\frac{17}{16}\sec w\implies du=\sqrt\frac{17}{16}\sec w\tan w\,dw$ \begin{align}\int\frac{\sec w\tan w\,dw}{\sec w\cdot\sqrt{17}\tan w }&=\frac w{\sqrt{17}}+C\\&=\frac1{\sqrt{17}}\sec^{-1}\left(\sqrt\frac{16}{17}x^3\right)\\&=\frac{1}{\sqrt{17}}\tan^{-1}\sqrt{\frac{16}{17}x^6-1}\end{align}
Here in the last step, we use: \begin{align}\sec^{-1}(\alpha)&=\beta\\\alpha&=\sec\beta\\\alpha^2&=\tan^2\beta+1\\\tan\beta&=\sqrt{\alpha^2-1}\\\sec^{-1}(\alpha)=\beta&=\tan^{-1}\left(\sqrt{\alpha^2-1}\right)\end{align}