# Evaluate $\int \frac{\sqrt{64x^2-256}}{x}\,dx$

$$\int \frac{\sqrt{64x^2-256}}{x}\,dx$$

Image.

I've tried this problem multiple times and cant seem to find where I made a mistake. If someone could please help explain where I went wrong I would really appreciate it

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$\newcommand{\dd}{\; \mathrm{d}}\int \frac{\sqrt{64x^2-256}}x \dd x \to \int \frac{\sqrt{64(x^2-4)}}x \dd x \to \int \frac{8\sqrt{x^2-4}}x \dd x$

Use $x=a\sec\theta$, $\dd x=a\sec\theta \tan\theta \dd \theta$.

$a=2$ $\to$ $x=2\sec\theta$, $\dd x=2\sec\theta \tan\theta \dd \theta$.

$=\int \frac{8\sqrt{4\sec^2\theta-4}}{2\sec\theta}(2\sec\theta\tan\theta) \dd \theta \to \int \frac{8\sqrt{4(\sec^2\theta-1)}}{2\sec\theta}(2\sec\theta\tan\theta) \dd \theta$

$=\int \frac{8\sqrt{4\tan^2\theta}}{2\sec\theta}(2\sec\theta\tan\theta) \dd \theta \to \int \frac{8(2\tan\theta)}{2\sec\theta}(2\sec\theta\tan\theta) \dd \theta$

$=\int 16\tan^2\theta \dd \theta \to 16\int\tan^2\theta \dd \theta \to \underset{\text{trig. formula}}{\underbrace{16(\theta+\tan\theta)+C}}$

$\Rightarrow 16(\tan\theta-\theta)+C = 16\tan\theta-16\theta$

$x=2\sec\theta$, $\sec\theta= \frac x2$

$\boxed{16\tan\left(\frac{\sqrt{x^2-4}}2\right) -16\sec^{-1}\left(\frac x2\right)+C}$

• Poor quality images of handwritten work are usually received poorly on Math SE. Try to type up all your work in your posts. Formatting tips here. – Em. Jul 13 '16 at 4:07
• The substitution $x=\frac{2}{\sin\theta}$ gives an easy integral. – Jack D'Aurizio Jul 13 '16 at 4:09
• I have tried to edit your attempt based on the images you posted. This should make it easier for you to edit it further into the form which really reflects what you want to say. – Martin Sleziak Jul 27 '16 at 8:51

Since I am almost blind, I have a lot of problems reading the image.

Consider $$I=\int \frac{\sqrt{64 x^2-256}}{x}\,dx$$ What you apparently did is $x=2\sec(t)$, $dx=2 \tan (t) \sec (t)$ which make $$I=\int \tan (t) \sqrt{256 \sec ^2(t)-256}\,dt=16\int \tan (t) \sqrt{\tan ^2(t)}\,dt=16\int \tan^2 (t) \,dt$$ $$I=16\int (1+\tan^2(t)-1)\,dt=16 (\tan (t)-t)$$

• Hi i did this and found t to equal arcsec(x/2) but when I plugged this into the equation it says that the answer isn't correct. – cchang Jul 13 '16 at 4:38
• If you properly replace $t$, you should arrive to $2 \sqrt{x^2-64}+16 \tan ^{-1}\left(\frac{8}{\sqrt{x^2-64}}\right)$ – Claude Leibovici Jul 13 '16 at 4:45

$$\dfrac{\sqrt{64x^2-256}}x=8x\cdot\dfrac{\sqrt{x^2-4}}{x^2}$$

Let $$\sqrt{x^2-4}=y\implies x^2-4=y^2\implies x\ dx= y\ dy$$

$$\int\dfrac{\sqrt{64x^2-256}}xdx=8\int\dfrac{y^2dy}{y^2+4}=8\int dy-32\int\dfrac{dy}{y^2+4}=?$$

• @user376343,Thanks for your feedback – lab bhattacharjee Nov 30 '18 at 17:07


With the sub$\ds{\ldots\ t \equiv x - \root{x^{2} - 4}\ \imp\ x = {t^{2} + 4 \over 2t}}$:

\begin{align} &\color{#f00}{\int{\root{64x^{2} - 256} \over x}\,\dd x} = 8\int{\root{x^{2} - 4} \over x}\,\dd x = 8\int\pars{{8 \over t^{2} + 4} - {16 \over t^{2}} + 3}\,\dd t \\[3mm] = &\ 32\arctan\pars{t \over 2} + {128 \over t} + 24t \\[3mm] = &\ 32\arctan\pars{x - \root{x^{2} - 4} \over 2} + {128 \over x - \root{x^{2} - 4}} + 24\pars{x - \root{x^{2} - 4}} \\[3mm] = &\ 32\arctan\pars{x - \root{x^{2} - 4} \over 2} + 32\pars{x + \root{x^{2} - 4}} + 24\pars{x - \root{x^{2} - 4}} \\[3mm] = &\ \color{#f00}{32\arctan\pars{x - \root{x^{2} - 4} \over 2} + 56x + 8\root{x^{2} - 4}} + \pars{~\mbox{a constant}~} \end{align}