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I'm trying to integrate $$\int \frac{5-4x}{\sqrt{x^2-3x+2}}dx$$

I've tried u-substitution($u=x^2-3x+2$), but I can't seem to figure out how to make it work, so I've tried completing the square:

$$\:\int \frac{5-4x}{\sqrt{\left(x-\frac{3}{2}\right)^2-\left(\frac{1}{2}\right)^2}}dx$$

But I'm still stuck on where to go from there. I'm guessing I need to use the $\int \frac{du}{\sqrt{u^2-a^2}}$ formula, but I can't figure out what to do with $5-4x$.

Integreek
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Hx0
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5 Answers5

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Write $5-4x=-1-4\left(x-\dfrac32\right)$

So, we have $$-4\int\dfrac{\left(x-\dfrac32\right)dx}{\sqrt{\left(x-\dfrac32\right)^2-\left(\dfrac12\right)^2}}-\int\dfrac{dx}{\sqrt{\left(x-\dfrac32\right)^2-\left(\dfrac12\right)^2}}$$

For the first integral, set $\left(x-\dfrac32\right)^2-\left(\dfrac12\right)^2=u$

19aksh
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lab bhattacharjee
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Hint-

$5-4x=\alpha(2x-3)+\gamma$

You can find both $\alpha, \gamma$ by comparing terms on both sides and then express it in the form of the derivative and use the standard formula for the denominator. $$\int\frac{-2(2x-3)-1}{\sqrt{x^2-3x+2}}dx=-2\int \frac{2x-3}{\sqrt{x^2-3x+2}}dx \ -\int \frac{1}{\sqrt{x^2-3x+2}}dx$$

Vyom Yadav
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Directly substitute $u^2=x^2-3x+2\implies (x-\dfrac32)\,dx=u\,du$.

Note that $(5-4x+1)-1=\color{blue}{(6-4x)}-1=\color{blue}{-4(x-\dfrac32)}-1.$ Now, you can separate the integral and use $\displaystyle\int \frac{du}{\sqrt{u^2-a^2}}$ for the second half.

poyea
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By completing the square in the denominator, one can see that it would be feasible to perform the substitution $x=\frac{3+\cosh t}2$:

$$\mathcal I=\int \frac{5-4x}{\sqrt{x^2-3x+2}}\mathrm dx$$ $$=\int-2\cosh t-1\mathrm dt$$ $$=-2\sinh t-t+C$$

Now, just do the back substitution.

Integreek
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0

Write the numerator as

$$-4x+5=-3(x-1)-(x-2)$$

Since the domain of the integrand is $(-\infty,1)\cup(2,\infty)$, we make $2$ cases:

$\textbf{Case 1, $x>2$:}$ $$\begin{align}\int\frac{5-4x}{\sqrt{x^2-3x+2}}&=-3\int\frac{\sqrt{x-1}}{\sqrt{x-2}}\mathrm dx-\int\frac{\sqrt{x-2}}{\sqrt{x-1}}\mathrm dx\\&=-6\int\sqrt{\left(\sqrt{x-2}\right)^2+1}\text{ }\mathrm d\left(\sqrt{x-2}\right)-2\int\sqrt{\left(\sqrt{x-1}\right)^2-1}\text{ }\mathrm d\left(\sqrt{x-1}\right)\end{align}$$

$\textbf{Case 2, $x<1$:}$ $$\begin{align}\int\frac{5-4x}{\sqrt{x^2-3x+2}}&=3\int\frac{\sqrt{1-x}}{\sqrt{2-x}}\mathrm dx+\int\frac{\sqrt{2-x}}{\sqrt{1-x}}\mathrm dx\\&=-6\int\sqrt{\left(\sqrt{2-x}\right)^2-1}\text{ }\mathrm d\left(\sqrt{2-x}\right)-2\int\sqrt{\left(\sqrt{1-x}\right)^2+1}\text{ }\mathrm d\left(\sqrt{1-x}\right)\end{align}$$

Integreek
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