How to integrate the product of two or more polynomials raised to some powers, not necessarily integral

Question

This question is inspired by my own answer to a question which I tried to answer and got stuck at one point.

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The question was:

HI DARLING.

USE MY ATM CARD, TAKE ANY AMOUNT OUT, GO SHOPPING AND TAKE YOUR FRIENDS FOR LUNCH.

PIN CODE: $\displaystyle \int_{0}^{1} \frac{3x^3 - x^2 + 2x - 4}{\sqrt{x^2 - 3x + 2}} \, dx $

I LOVE YOU HONEY.

Anyone knows? Are we gonna get an integer number?

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My attempt:

Does this help?

$$\frac{3x^3-x^2+2x-4}{x-1}=3x^2+2x+4$$ (long division) \begin{align*} I&=\int\frac{3x^3-x^2+2x-4}{[(x-1)(x-2)]^{1/2}} dx = \\ &=\int\frac{(3x^2+2x+4)(x-1)^{1/2}}{(x-2)^{1/2}} dx = \\ &=\int 3(u^4-4u^2-4)(u^2+1)^{1/2}du \times 2 \end{align*} after the substitution \begin{gather*} (x-2)^{1/2}=u\\ du=\frac1{2(x-2)^{1/2}}dx\\ u^2=x-2\\ (x-1)^{1/2}=(u^2+1)^{1/2} \end{gather*}

Update: This may help us proceed.

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I tried to proceed:

$$6\int (u^4-4u^2-4)(u^2+1)^{1/2} du = 6\int ((t-3)^2-8)t \frac{dt}{2u}$$ after $u^2+1=t$ and $dt=2udu$ \begin{align*} u^4-4u^2-4 &= (u^2+1)^2-(6u^2+5) \\ &= (u^2+1)^2-6(u^2+1)+1 \\ &= ((u^2+1)-3)^2-8 \end{align*}

I wonder whether this question can be solved from here?

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Update:

This has been getting a lot of views, and I think most people came for the sort of problem mentioned in the title (where I got stuck) rather than the original problem itself.

Keepin this in mind, I'm reopening the question and here's the kind of answers I expect — Solutions to the original problem are good, but I'd prefer solutions that continue from the part where I got stuck — the polynomial in $u$ — that's the sort of problem mentioned in the title.

Answer 1

Looking at your previous deleted post, one answer suggested to use Euler subtitution $$\sqrt{x^2-3x+2}=t+x\implies x=\frac{2-t^2}{2t+3}\implies dx=-\frac{2 (t+1) (t+2)}{(2 t+3)^2}\,dt$$ Replacing, we arrive to $$\frac{3x^3 - x^2 + 2x - 4}{\sqrt{x^2 - 3x + 2}}=\frac{2 (t+1)^2 \left(3 t^4-4 t^3-2 t^2+56 t+60\right)}{(2 t+3)^4}$$ Now let $2t+3=u$ to make the integrand $$\frac{3 u^2}{64}-\frac{25 u}{32}+\frac{317}{64}-\frac{135}{16 u}+\frac{317}{64 u^2}-\frac{25}{32 u^3}+\frac{3}{64 u^4}$$ and the antiderivative $$\frac{u^3}{64}-\frac{25 u^2}{64}+\frac{317 u}{64}-\frac{135}{16} \log \left({u}\right)-\frac{317}{64 u}+\frac{25}{64 u^2}-\frac{1}{64 u^3}$$ For $t$, the bounds were $(\sqrt 2,-1)$; so, for $u$, they are $(2\sqrt 2+3,1)$ giving as a result $$ \int_{0}^{1} \frac{3x^3 - x^2 + 2x - 4}{\sqrt{x^2 - 3x + 2}} \, dx=\frac{135}{16} \log \left(3+2 \sqrt{2}\right)-\frac{101}{4 \sqrt{2}}\approx -2.98127$$

Answer 2

Alternative method: $ \def\lfrac#1#2{{\large\frac{#1}{#2}}} $

Express the integrand in the form $\lfrac{(2ax+b)·(x^2-3x+2)+(2cx+d)}{\sqrt{x^2-3x+2}}$ for some constants $a,b,c,d$.

Then split it into $( a(2x-3) + (3a+b) ) · \sqrt{x^2-3x+2} + \lfrac{c(2x-3)+(3c+d)}{\sqrt{x^2-3x+2}}$, so that as a sum of four terms the first and third have obvious antiderivatives. The other two terms can be solved by standard techniques.

Answer 3

$\require{begingroup}\begingroup$This should help to get closer to the final result (if you want to calculate this manually): $$\newcommand{\dd}{\; \mathrm{d}} I=\int_0^1 \frac{3x^3 - x^2 + 2x - 4}{\sqrt{x^2 - 3x + 2}} \dd x = \int_0^1 \frac{3x^3 - x^2 + 2x - 4}{\sqrt{(x-\frac32)^2 -\frac14}} \dd x$$ It will take some computing but we can get that $3x^3 - x^2 + 2x - 4 = 3(x-\frac32)^3+\frac{25}2(x-\frac32)^2+\frac{77}4(x-\frac32)+\frac{55}8$.

So we have $$I= \int_{-3/2}^{-1/2} \frac{3u^3+\frac{25}2u^2+\frac{77}4u+\frac{55}8}{\sqrt{u^2-\frac14}} \dd u = \begin{vmatrix} t=2u & u=\frac t2 \\ \dd t=2\dd u & \dd u = \frac12\dd t \end{vmatrix} = \frac12 \int_{-3}^{-1} \frac{\frac{3t^3}8+\frac{25}8t^2+\frac{77}8t+\frac{55}8}{\sqrt{\frac{t^2}4-\frac14}} \dd t = \frac18 \int_{-3}^{-1} \frac{3t^3+25t^2+77t+55}{\sqrt{t^2-1}} \dd t = \frac18 \int_{-3}^{-1} \frac{3t(t^2-1)+25(t^2-1)+80t+80}{\sqrt{t^2-1}} \dd t = \frac18 \int_{-3}^{-1} (3t+25)\sqrt{t^2-1} +80 \frac{t+1}{\sqrt{t^2-1}} \dd t $$

You can check that Wolfram Alpha returns the same value for the original integral and this integral. (To be honest, I am not sure how I am supposed to get a PIN number from the result.)

Now you could divide up this into separate integrals which should be not too difficult:

• How to calculate this integral with square roots: $\int\frac{ \sqrt{x+1} }{ \sqrt{ x-1 }} \, dx$
• Indefinite integral of $\int\sqrt{x^2-1}dx$
• For $\int t\sqrt{t^2-1} \dd t$ the substitution $s=t^2-1$ seems reasonable.

$\endgroup$

Answer 4

Noticing that $3 x^3-x^2+2 x-4=-(1-x)\left(3 x^2+2 x+4\right)$ and $x^2-3 x+2=(1-x)(2-x)$, we transform the integral into $$ I=-\int_0^1 \sqrt{\frac{1-x}{2-x}}\left(3 x^2+2 x+4\right) d x $$ Let $t^2=\frac{1-x}{2-x}$, then $x=\frac{2 t^2-1}{t^2-1}$ and $d x=-\frac{2 t}{\left(t^2-1\right)^2} d t$, which changes the integral into $$ I=-2 \int_0^{\frac{1}{\sqrt{2}}} \frac{t^2\left(20 t^4-26 t^2+9\right)}{\left(t^2-1\right)^4} d t $$ Resolving the integrand into partial fractions as $$ \frac{t^2\left(20 t^4-26 t^2+9\right)}{\left(t^2-1\right)^4} = {-\frac{135}{32} \cdot \frac{1}{t+1}+\frac{185}{32} \cdot \frac{1}{(t+1)^2}-\frac{7}{4} \cdot \frac{1}{(t+1)^3}+\frac{3}{16} \cdot \frac{1}{(t+1)^4}}+\frac{135}{32} \cdot \frac{1}{t-1}+\frac{185}{32} \cdot \frac{1}{(t-1)^2}+\frac{7}{4} \cdot \frac{1}{(t-1)^3}+\frac{3}{16}\cdot \frac{1}{(t-1)^4}$$ Integrating it from $0$ to $\frac{1}{\sqrt 2} $ yields $$ \begin{aligned} I=&-2 \left[-\frac{135}{32} \ln |t+1|-\frac{185}{32(t+1)}+\frac{7}{8(t+1)^2}-\frac{1}{16(t+1)^3} +\frac{135}{32} \ln |t-1|\right.\\ &\left.-\frac{185}{32(t-1)}-\frac{7}{8(t-1)^2}-\frac{1}{16(t-1)^3}\right]_0^{\frac{1}{\sqrt{2}}}\\=& \frac{135}{16} \ln (3+2 \sqrt{2})-\frac{101 \sqrt{2}}{8} \end{aligned} $$

Answer 5

Noticing that $ 3 x^3-x^2+2 x-4=\frac{3}{2}(2 x-3)\left(x^2-3 x+2\right)+\frac{5}{2}\left(5 x^2-7 x+2\right),$ we have $$ I=\frac{3}{2} \underbrace{\int_0^1(2 x-3) \sqrt{x^2-3 x+2} d x}_J+\frac{5}{2} \underbrace{\int_0^1 \frac{5 x^2-7 x+2}{\sqrt{x^2-3 x+2}} d x}_K $$$$ \begin{aligned} J =\frac{3}{2} \int_0^1 \sqrt{x^2-3 x+2} d\left(x^2-3 x+2\right) =\left[\left(x^2-3 x+2\right)^{\frac{3}{2}}\right]_0^1=-2 \sqrt{2} \end{aligned} $$ Again, decompose the numerator of the integrand of $K$ into 3 parts as $$ 5 x^2-7 x+2=5\left(x^2-3 x+2\right)+4(2 x-3)+4, $$ we get $$ \begin{aligned} K&=5 \int_0^1 \sqrt{x^2-3 x+2} d x+4 \int_0^1 \frac{d\left(x^2-3 x+2\right)}{\sqrt{x^2-3 x+2}} +4 \int \frac{d x}{\sqrt{x^2-3 x+2}}\\&= 5 \cdot \frac{1}{8}(6 \sqrt{2}-\ln (3+2 \sqrt{2}))-8 \sqrt{2}+4 \ln (3+2 \sqrt{2}) \cdots(*)\\&= -\frac{17}{4} \sqrt{2}+\frac{27}{8} \ln (3+2 \sqrt{2}) \end{aligned} $$ where $(*)$ comes from my post.

Plugging back yields $$ \boxed{I=\frac{135}{16} \ln (3+2 \sqrt{2})-\frac{101}{8} \sqrt{2}} $$