Evaluation of the integral $\int_{-6}^{-3}\frac{\sqrt{x^2-9}}{x}$

Question

How to evaluate the following integral?

I have tried the following things but I have no idea to continue after the last step. Moverover, the integral seems wrong when compared with the ans from WolframAlpha app.

$$=\int_{-6}^{-3}{\sqrt{x^2-9}\over x}\,dx$$

Substitute $u=\sqrt{x^2-9}\implies du ={x\over \sqrt{x^2-9}}$

$$=\int_{x=-6}^{x=-3}{\sqrt{x^2-9}\over x}\cdot{\sqrt{x^2-9}\over x}\,dx$$

$$=\int_{x-6}^{x=-3}{x^2-9 \over x^2}\,du$$

$$=\int_{x=-6}^{x=-3}{u^2\,du \over u^2 + 9}$$

$$=\int_{x=-6}^{x=-3}{du\over 1 + \left(\frac3{u}\right)^2}$$

$$=\left[\tan^{-1}\left({3\over \sqrt{x^2-9}}\right)\right]_{-6}^{-3}$$

$$=\pi/2$$

While WolframAlpha gives $\pi -3\sqrt{3}$

This is a link to my original work

Thank you for your attention.

Answer 1

Your antiderivative is incorrect. You can see this by differentiating $\arctan (\frac{3}{u})$; you will not get the original integrand back. I recommend performing long division on $\frac{u^2}{u^2+9}$ to receive $1-\frac{9}{u^2+9}$. Proceed from there and you will arrive at the correct answer.

And you should also change the bounds of integration appropriately.

Answer 2

Let us see. We have to evaluate: $$I=\int_{-6}^{-3}\frac{\sqrt{x^2-9}}{x}\mathrm{d}x.$$ You suggest substituting $u=\sqrt{x^2-9}$, thus getting $\mathrm{d}u=\frac{x}{\sqrt{x^2-9}}\mathrm{d}x\implies\mathrm{d}u=\frac{u}{-\sqrt{u^2+9}}\mathrm{d}u$. The minus is due to the fact that we are integrating over $[-6,-3]$ where $x<0$ and the root would be positive. Plug that in and we get: $$I=\int_{\sqrt{27}}^0\frac{u}{-\sqrt{u^2+9}}\frac{u}{\sqrt{u^2+9}}\mathrm{d}u=\int_{3\sqrt3}^0\frac{u^2}{u^2+9}\mathrm{d}u=u\Big|_{3\sqrt3}^0-\int_{3\sqrt3}^0\frac{9}{u^2+9}\mathrm{d}u.$$ The first piece evaluates to $-3\sqrt3$. For the second piece we let $t=\frac{u}{3}$, whence $\mathrm{d}t=\frac{\mathrm{d}u}{3}$. That bit therefore becomes: $$-\int_{3\sqrt3}^0\frac{-9}{u^2+9}\mathrm{d}u=-3\int_{\sqrt3}^0\frac{1}{t^2+1}\mathrm{d}t=-3\arctan(t)\Big|_{\sqrt3}^0=3\arctan(\sqrt3).$$ So the integral finally becomes: $$I=-3\sqrt3+3\arctan(\sqrt3)=-3\sqrt3+3\frac{\pi}{3}=\pi-3\sqrt3.$$ Your mistake was when you said: $$\int\frac{\mathrm{d}u}{1+\left(\frac3u\right)^2}=\tan^{-1}\left(\frac3u\right).$$ The integral of $\frac{1}{1+x^2}\mathrm{d}x$ is the arctangent, but there you don't have that, to have that you'd need $\mathrm{d}\!\left(\frac3u\right)$, but you have $\mathrm{d}u$, which is not the same.

Answer 3

According to me answer is

$ 3√3 - \pi$

As According to Wolfram Alpha answer is

$\pi - 3√3$.

So I doubt I might have made a mistake but I am not able it find it.