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I've got one integration question which I first felt was not a hard nut to crack. But, as I proceeded, difficulties arose. This is the one:

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

I went ahead simplifying the two expressions and ultimately I reached this step:

$\displaystyle\int_{0}^{1}{\sqrt{\frac{x-1}{x-2}} \ (3x^2 + 2x + 4) \ dx}$

I don't now what to do now, had I followed the correct pathway? Is there any other simpler method?

Archer
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Simar
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2 Answers2

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You can try a hyperbolic substitution: $$\sqrt{x^2-3x+2}=\sqrt{\Bigl(x-\frac{3\strut}2\Bigr)^{2}-\frac94+2}=\frac12\sqrt{(2x-3)^2-1},$$ so you can set, for $t\ge 0$, $$2x-3=\cosh t\iff x=\frac{\cosh t+3}2,\qquad\mathrm dx=\frac12\sinh t\,\mathrm dt.$$ and the denominator of the integrand becomes $\;\frac12\sinh t$.

After you've simplified, you should obtain the integral of a cubic polynomial in $\cosh t$.

Bernard
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  • I completely understood your initial step, but when you came to $\cosh(?)$, I became a bit upset because I don't know why you did this, what was the reason? Method? as I'm a grade 12 student! – Simar Jul 20 '18 at 15:37
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    The idea is to remove the square root using a functional identity. For instance, in high school, it's quite common to remove the square root in $\sqrt'1-x^2}$ setting $x=\sin t$ ($-\pi/2\le t\le\pi/2$), because of Pythagoras' identity $\sin^2t+\cos^2t=1$, s that $:\sqrt{1-x^2}=\sqrt{1-\sin^2t}=\sqrt{\cos^2t}=|\cos t|=\cos t$ since $t$ lies between $-\pi/2$ and $\pi/ 2$. Here I simply used the analogous identity for hyperbolic functions: $\cosh^2t-\sinh^2t=1$. – Bernard Jul 20 '18 at 15:45
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    Oh my God, yes, I got this now, after trying and trying and trying!!! THANKS!, The more I appreciate your efforts, the less my appreciation goes..... – Simar Jul 20 '18 at 15:55
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It can be seen easily that for $H\left( x \right)=\sqrt{{{x}^{2}}-3x+2}$ $$\begin{align} & {{\left( {{x}^{2}}H\left( x \right) \right)}^{\prime }}=\frac{6{{x}^{3}}-15{{x}^{2}}+8x}{2H\left( x \right)}, \\ & {{\left( xH\left( x \right) \right)}^{\prime }}=\frac{4{{x}^{2}}-9x+4}{2H\left( x \right)}, \\ & {{\left( H\left( x \right) \right)}^{\prime }}=\frac{2x-3}{2H\left( x \right)} \\ \end{align}$$ This suggests that there exists $a,b,c\ and\ d$ $$\frac{3{{x}^{3}}-{{x}^{2}}+2x-4}{H\left( x \right)}=a{{\left( {{x}^{2}}H\left( x \right) \right)}^{\prime }}+b{{\left( xH\left( x \right) \right)}^{\prime }}+c{{\left( H\left( x \right) \right)}^{\prime }}+\frac{d}{H\left( x \right)}$$ Hence $$\frac{3{{x}^{3}}-{{x}^{2}}+2x-4}{H\left( x \right)}=a\frac{6{{x}^{3}}-15{{x}^{2}}+8x}{2H\left( x \right)}+b\frac{4{{x}^{2}}-9x+4}{2H\left( x \right)}+c\frac{2x-3}{2H\left( x \right)}+\frac{d}{H\left( x \right)}$$ Or $$3{{x}^{3}}-{{x}^{2}}+2x-4=\frac{a}{2}\left( 6{{x}^{3}}-15{{x}^{2}}+8x \right)+\frac{b}{2}\left( 4{{x}^{2}}-9x+4 \right)+\frac{c}{2}\left( 2x-3 \right)+d$$ Equating the coefficients yields $$\begin{align} & a=1 \\ & -15a+4b=-2 \\ & 8a-9b+2c=4 \\ & 4b-3c+2d=-8 \\ \end{align}$$ and we get $$a=1,b=\frac{13}{4},c=\frac{101}{16},d=\frac{135}{16}$$ Finally $$\int{\frac{3{{x}^{3}}-{{x}^{2}}+2x-4}{H\left( x \right)}dx={{x}^{2}}H\left( x \right)}+\frac{13}{4}xH\left( x \right)+\frac{101}{16}H\left( x \right)+\frac{135}{16}\int{\frac{dx}{H\left( x \right)}}$$