Integrate $\int \frac{x^2}{(x^2+2x+2) \sqrt{x-1}} dx$

Question

Find value of $$\int\frac{x^2}{(x^2+2x+2) \sqrt{x-1}} dx$$

So try $\sqrt{x-1} = u$ to get $dx = 2udu$ $$\int \frac{2u(u^2+1)^2 du}{ ((u^2+2)^2+1)u} = 2\int \frac{u^4 + 2u^2+1}{u^4+4u^2+5} = 2\left(\int1 \cdot du - \int\frac{2u^2 +4}{u^4+4u^2+5}\right)\\ = 2( u - J)$$

To evaluat $J$, I divide by $u^2$ on numerator and denominator to get $\frac{2+4u^{-2}}{u^2+4+5u^{-2}}$. Now I try to split into numerator into $$2+4u^{-2} = a (1-\sqrt{5}u^{-2})+b(1+\sqrt{5}u^{-2})$$ to get $a = 1 - \frac{2}{\sqrt{5}} $ and $b = 1+\frac{2}{\sqrt{5}}$.

$$J = \int \frac{a\, d(u+\sqrt{5} u^{-1})}{(u + \sqrt{5} u^{-1})^2-(2\sqrt{5}-4)} du + \int \frac{b\, d(u-\sqrt{5} u^{-1})}{(u - \sqrt{5} u^{-1})^2+(2\sqrt{5}+4)}$$

So this lead to final answer

$$2\left(u - \left( \frac{1-\frac{2}{\sqrt{5}}}{2 \sqrt{2\sqrt{5}-4 }} \ln\left|\frac{ \sqrt{2\sqrt{5}-4 }-u-\frac{\sqrt{5}}{u}}{ \sqrt{2\sqrt{5}-4 }+u+\frac{\sqrt{5}}{u}}\right| + \frac{1+\frac{2}{\sqrt{5}}}{ \sqrt{2\sqrt{5}+4 }} \arctan \left(\frac{u - \frac{\sqrt 5}{u}}{\sqrt{2\sqrt{5}+4 }}\right)\right)\right)+C$$

our miss ask to check all our integration, how can i even check its validity o differentiation will make me mad.

Can you give simple effort to find integral simple forms.

Note I edited question because I make a typo. I changed $x^2+2 $ to $x^2$. Even if it was $x^2+2$ in numerator, only $a,b$ change in my answer! So I changed back as we can take it either way! I hope you understand.

Answer 1

Let us start with $$\frac{2u^2 +4}{u^4+4u^2+5}=\frac{2u^2 +4}{(u^2-a)(u^2-b)}$$ where $a=-2-i$ and $b=-2+i$.

Now, using partial fraction decomposition, $$\frac{2u^2 +4}{(u^2-a)(u^2-b)}=\frac{2 (a+2)}{(a-b) \left(u^2-a\right)}-\frac{2 (b+2)}{(a-b) \left(u^2-b\right)}$$

Just continue and, at the end, play with the complex numbers.

Answer 2

You messed up when you did your substitution of $u$ into the numerator. It should be $2u((u^2+1)^2+2) \; \mathrm{d}u$, not $2u(u^2+1)^2 \; \mathrm{d}u$.

Anyways, here's a cut down solution (cut down because it's sadistic):

We are required to find the value of $$I=\displaystyle \int \dfrac{x^2+2}{(x^2+2x+2)\sqrt{x-1}} \; \mathrm{d}x$$

We let $u=\sqrt{x-1}$, so $\mathrm{d}x = 2u \; \mathrm{d}u$. Substituting, we get $$\begin{aligned} I&=\displaystyle \int \dfrac{(u^2+1)^2+2}{((u^2+1)^2+2(u^2+1)+2))u} 2u \; \mathrm{d}u\\ &=2 \displaystyle \int \dfrac{(u^2+1)^2+2}{u^4+4u^2+5} \; \mathrm{d}u\\ &=2 \left (\displaystyle \int 1 \; \mathrm{d}u - \displaystyle \int \dfrac{2u^2+2}{u^4+4u^2+5} \; \mathrm{d}u \right )\\ &= 2 \left (\displaystyle \int 1 \; \mathrm{d}u - 2 \displaystyle \int \dfrac{u^2+1}{u^4+4u^2+5} \; \mathrm{d}u \right ) \end{aligned}$$

Now, I'm just gonna let $$\begin{aligned} J&=\displaystyle \int \dfrac{u^2+1}{u^4+4u^2+5} \; \mathrm{d}u\\ &= \dfrac{1}{2\sqrt{10\sqrt{5}-20}} \displaystyle \int \dfrac{(\sqrt{5}-1)u+\sqrt{2\sqrt{5}-4}}{u^2-\sqrt{2\sqrt{5}-4}u+\sqrt{5}} \; \mathrm{d}u - \dfrac{1}{2\sqrt{10\sqrt{5}-20}} \displaystyle \int \dfrac{(\sqrt{5}-1)u-\sqrt{2\sqrt{5}-4}}{u^2+\sqrt{2\sqrt{5}-4}u+\sqrt{5}} \; \mathrm{d}u \end{aligned}$$

(Here, I skipped over the method of partial fraction decomposition because it's painful).

Now, these integrals are standard (linear on quadratic), so you should be able to solve them yourself, but the final (ugly) result is

$$J=\dfrac{\sqrt{5}-1}{4\sqrt{10\sqrt{5}-20}} (\ln (u^2+\sqrt{2\sqrt{5}-4}u+\sqrt{5})+\ln(u^2-\sqrt{2\sqrt{5}-4}u+\sqrt{5}))+\dfrac{\sqrt{\sqrt{5}-2}(\sqrt{5}+1)}{2\sqrt{\sqrt{5}+2}\sqrt{10\sqrt{5}-20}} \left ( \tan^{-1} \left (\dfrac{\sqrt{2}u+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+2}} \right ) + \tan^{-1} \left (\dfrac{\sqrt{2}u-\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+2}} \right )\right)$$

Substituting that in for $I$ gives the final expression in $x$ (which I will not show how to derive because it is long) is

$$\begin{aligned} I&=\dfrac{1}{\sqrt{10}} \left ( \sqrt{\sqrt{5}-2} \left ( \sqrt{5}+3 \right ) \ln \left | \dfrac{x+\sqrt{2}\sqrt{\sqrt{5}-2}\sqrt{x-1}+\sqrt{5}-1}{x-\sqrt{2}\sqrt{\sqrt{5}-2}\sqrt{x-1}+\sqrt{5}-1} \right | + 2 \left ( \sqrt{5}-3 \right ) \sqrt{\sqrt{5}+2} \tan^{-1} \left ( \left ( \sqrt{5}-2 \right )\sqrt{\sqrt{5}+2} \left ( \sqrt{2}\sqrt{x-1}+\sqrt{\sqrt{5}-2} \right ) \right ) + \tan^{-1} \left ( \left ( \sqrt{5}-2 \right ) \sqrt{\sqrt{5}+2} \left ( \sqrt{2}\sqrt{x-1} -\sqrt{\sqrt{5}-2} \right ) \right ) \right )+2\sqrt{x-1}+C \end{aligned}$$

Also, this is an ugly integral in all regards, and should never be given in a test. To verify integrals, just use https://www.integral-calculator.com/. It calculates integrals and shows the steps if you need it.

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EDIT: Oh no, you edited and I never realised. Just follow what I did, but just do some tweaks.