Improper Integration Technique

Question

What would be the best techniques to start here?

$$\int_1^2 \frac{(x+2)^2}{\sqrt{1-1/x^2}} \, dx $$

I have tried substitution but it seems to take a while, I was thinking $$u=1/x$$ but still not sure? Any Help it greatly appreciated!

Answer 1

You have $$I= \int_1^2 \frac{(x+2)^2}{\sqrt{1- \frac{1}{x^2}}} \mathrm{dx} = \int_1^2 \frac{x(x+2)^2}{\sqrt{x^2-1}} \mathrm{dx}$$

Now, the substitution $y= \mathrm{argch}(x)$ gives you a closed form.

Answer 2

Well, we can do the following:

$$\mathcal{I}_\text{n}:=\int_1^\text{n}\frac{\left(2+x\right)^2}{\sqrt{1-\frac{1}{x^2}}}\space\text{d}x\tag1$$

Let's substitute $x=\sec\left(\text{u}\right)$, so we get:

$$\mathcal{I}_\text{n}=\int_0^{\text{arcsec}\left(\text{n}\right)}\sec^2\left(\text{u}\right)\cdot\left(2+\sec\left(\text{u}\right)\right)^2\space\text{d}\text{u}=$$ $$\int_0^{\text{arcsec}\left(\text{n}\right)}\left(\sec^4\left(\text{u}\right)+4\cdot\sec^3\left(\text{u}\right)+4\cdot\sec^2\left(\text{u}\right)\right)\space\text{d}\text{u}=$$ $$\int_0^{\text{arcsec}\left(\text{n}\right)}\sec^4\left(\text{u}\right)\space\text{d}\text{u}+4\cdot\int_0^{\text{arcsec}\left(\text{n}\right)}\sec^3\left(\text{u}\right)\space\text{d}\text{u}+4\cdot\int_0^{\text{arcsec}\left(\text{n}\right)}\sec^2\left(\text{u}\right)\space\text{d}\text{u}\tag2$$

Now, using the reduction formula:

$$\int\sec^\text{p}\left(x\right)\space\text{d}x=\frac{\sin\left(x\right)\cdot\sec^{\text{p}-1}\left(x\right)}{\text{p}-1}+\frac{\text{p}-2}{\text{p}-1}\cdot\int\sec^{\text{p}-2}\left(x\right)\space\text{d}x\tag3$$

So:

• When $\text{p}=4$: $$\int_0^{\text{arcsec}\left(\text{n}\right)}\sec^4\left(x\right)\space\text{d}x=\left[\frac{\sin\left(x\right)\cdot\sec^{4-1}\left(x\right)}{4-1}\right]_0^{\text{arcsec}\left(\text{n}\right)}+\frac{4-2}{4-1}\cdot\int_0^{\text{arcsec}\left(\text{n}\right)}\sec^{4-2}\left(x\right)\space\text{d}x=$$ $$\frac{\text{n}^3}{3}\cdot\sqrt{1-\frac{1}{\text{n}^2}}+\frac{2}{3}\cdot\int_0^{\text{arcsec}\left(\text{n}\right)}\sec^2\left(x\right)\space\text{d}x\tag4$$
• When $\text{p}=2$: $$\int_0^{\text{arcsec}\left(\text{n}\right)}\sec^2\left(x\right)\space\text{d}x=\left[\frac{\sin\left(x\right)\cdot\sec^{2-1}\left(x\right)}{2-1}\right]_0^{\text{arcsec}\left(\text{n}\right)}+\frac{2-2}{2-1}\cdot\int_0^{\text{arcsec}\left(\text{n}\right)}\sec^{2-2}\left(x\right)\space\text{d}x=$$ $$\left[\frac{\sin\left(x\right)\cdot\sec^{2-1}\left(x\right)}{2-1}\right]_0^{\text{arcsec}\left(\text{n}\right)}=\text{n}\cdot\sqrt{1-\frac{1}{\text{n}^2}}\tag5$$

So, for now we get:

$$\mathcal{I}_\text{n}=\frac{\text{n}^3}{3}\cdot\sqrt{1-\frac{1}{\text{n}^2}}+\frac{2}{3}\cdot\text{n}\cdot\sqrt{1-\frac{1}{\text{n}^2}}+4\cdot\int_0^{\text{arcsec}\left(\text{n}\right)}\sec^3\left(\text{u}\right)\space\text{d}\text{u}+4\cdot\text{n}\cdot\sqrt{1-\frac{1}{\text{n}^2}}=$$ $$\frac{\text{n}\cdot\left(14+\text{n}^2\right)}{3}\cdot\sqrt{1-\frac{1}{\text{n}^2}}+4\cdot\int_0^{\text{arcsec}\left(\text{n}\right)}\sec^3\left(\text{u}\right)\space\text{d}\text{u}\tag6$$

Now, to finish it off we can just use the reduction formula ones again:

$$\int_0^{\text{arcsec}\left(\text{n}\right)}\sec^3\left(x\right)\space\text{d}x=\left[\frac{\sin\left(x\right)\cdot\sec^{3-1}\left(x\right)}{3-1}\right]_0^{\text{arcsec}\left(\text{n}\right)}+\frac{3-2}{3-1}\cdot\int_0^{\text{arcsec}\left(\text{n}\right)}\sec^{3-2}\left(x\right)\space\text{d}x=$$ $$\frac{\text{n}^2}{2}\cdot\sqrt{1-\frac{1}{\text{n}^2}}+\frac{1}{2}\cdot\int_0^{\text{arcsec}\left(\text{n}\right)}\sec\left(x\right)\space\text{d}x\tag7$$

Now, using this we now that:

$$\int_0^{\text{arcsec}\left(\text{n}\right)}\sec\left(x\right)\space\text{d}x=\ln\left|\tan\left(\frac{\text{arcsec}\left(\text{n}\right)}{2}+\frac{\pi}{4}\right)\right|\tag8$$

So, in the end we have:

$$\mathcal{I}_\text{n}=\frac{\text{n}\cdot\left(14+\text{n}^2\right)}{3}\cdot\sqrt{1-\frac{1}{\text{n}^2}}+4\cdot\left(\frac{\text{n}^2}{2}\cdot\sqrt{1-\frac{1}{\text{n}^2}}+\frac{1}{2}\cdot\ln\left|\tan\left(\frac{\text{arcsec}\left(\text{n}\right)}{2}+\frac{\pi}{4}\right)\right|\right)=$$ $$\frac{\text{n}\cdot\left(14+\text{n}\cdot\left(6+\text{n}\right)\right)}{3}\cdot\sqrt{1-\frac{1}{\text{n}^2}}+2\cdot\ln\left|\tan\left(\frac{\text{arcsec}\left(\text{n}\right)}{2}+\frac{\pi}{4}\right)\right|\tag9$$