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I want to integrate \begin{align} \int \frac{ev+f}{av^2 + bv +c} dv \end{align} can you give me some hints or detail procedure for this integral?

From mathematca, i have \begin{align} \frac{-\frac{2 (b e-2 a f) \text{ArcTan}\left[\frac{b+2 a v}{\sqrt{-b^2+4 a c}}\right]}{\sqrt{-b^2+4 a c}}+e \text{Log}[c+v (b+a v)]}{2 a} \end{align}

which seems uncomfortable for me..

phy_math
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  • Partial Fraction Decomposition. – Hosein Rahnama Apr 02 '16 at 13:38
  • partial fractions, maybe? Then complete the square of the quadratic term giving you the arctan, and the other giving you the Log part. – Sarvesh Ravichandran Iyer Apr 02 '16 at 13:39
  • The exact specific process depends on the signs of some of those constants. Do you have a specific example, or are you really supposed to integrate that general expression? If the latter, do you at least have restrictions on the constants? –  Apr 02 '16 at 13:48

3 Answers3

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$$ \int \frac{ev+f}{av^2 + bv +c} dv = \frac{e}{2a}\int \frac{d(av^2 + bv + c)}{av^2 + bv +c} + \left(f-\frac{be}{2a}\right)\int \frac{1}{av^2 + bv +c} dv $$ $$ = \frac{e}{2a} \ln(av^2+bv+c) + \left(f-\frac{be}{2a}\right)\int \frac{1}{(\sqrt{a}v+\frac{b}{2\sqrt{a}})^2 - \frac{b^2}{4a} +c} dv $$

$$ \int \frac{1}{(\sqrt{a}v+\frac{b}{2\sqrt{a}})^2 - \frac{b^2}{4a} +c} dv = \frac{1}{\sqrt{a}}\int \frac{d(\sqrt{a}v+\frac{b}{2\sqrt{a}})}{(\sqrt{a}v+\frac{b}{2\sqrt{a}})^2 - \frac{b^2}{4a} +c} $$

$$ c \geq \frac{b^2}{4a} \rightarrow k^2 = c - \frac{b^2}{4a} $$

$$ \frac{1}{\sqrt{a}}\int \frac{d(\sqrt{a}v+\frac{b}{2\sqrt{a}})}{(\sqrt{a}v+\frac{b}{2\sqrt{a}})^2 - \frac{b^2}{4a} +c} = \frac{1}{\sqrt{a}} \int \frac{du}{u^2+k^2} = \frac{1}{k\sqrt{a}}\arctan\left(\frac{u}{k}\right) = \\ \frac{1}{k\sqrt{a}}\arctan\left(\frac{\sqrt{a}v+\frac{b}{2\sqrt{a}}}{k}\right) = \\ \frac{1}{\sqrt{c - \frac{b^2}{4a}}\sqrt{a}}\arctan\left(\frac{\sqrt{a}v+\frac{b}{2\sqrt{a}}}{\sqrt{c - \frac{b^2}{4a}}}\right) $$ Then: $$ \int \frac{ev+f}{av^2 + bv +c} dv = \frac{e}{2a} \ln(av^2+bv+c) + \left(f-\frac{be}{2a}\right)[\frac{1}{\sqrt{c - \frac{b^2}{4a}}\sqrt{a}}\arctan\left(\frac{\sqrt{a}v+\frac{b}{2\sqrt{a}}}{\sqrt{c - \frac{b^2}{4a}}}\right)] + C $$ which is what you have. $$ c \leq \frac{b^2}{4a} \rightarrow k^2 = \frac{b^2}{4a} - c $$

$$ \frac{1}{\sqrt{a}}\int \frac{d(\sqrt{a}v+\frac{b}{2\sqrt{a}})}{(\sqrt{a}v+\frac{b}{2\sqrt{a}})^2 - \frac{b^2}{4a} +c} = \frac{1}{\sqrt{a}} \int \frac{du}{u^2-k^2} = \frac{1}{2k\sqrt{a}} \int {\frac{1}{u-k}-\frac{1}{u+k}} du = $$

$$ \frac{1}{2k\sqrt{a}}\ln (\frac{u-k}{u+k}) = \frac{1}{2\sqrt{\frac{b^2}{4a} - c}\sqrt{a}}\ln (\frac{\sqrt{a}v+\frac{b}{2\sqrt{a}}-\sqrt{\frac{b^2}{4a} - c}}{\sqrt{a}v+\frac{b}{2\sqrt{a}}+\sqrt{\frac{b^2}{4a} - c}}) $$ Then: $$ \int \frac{ev+f}{av^2 + bv +c} dv = \frac{e}{2a} \ln(av^2+bv+c) + \left(f-\frac{be}{2a}\right)[\frac{1}{2\sqrt{\frac{b^2}{4a} - c}\sqrt{a}}\ln (\frac{\sqrt{a}v+\frac{b}{2\sqrt{a}}-\sqrt{\frac{b^2}{4a} - c}}{\sqrt{a}v+\frac{b}{2\sqrt{a}}+\sqrt{\frac{b^2}{4a} - c}})] + C $$

Anonymous
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HINT:

Using Partial Fraction Decomposition let $ev+f=A\cdot\dfrac{d(av^2+bv+c)}{dv}+B$

Now $av^2+bv+c=\dfrac{(2av+b)^2+4ca-b^2}{4a}$

Now use Trigonometric substitution based on the sign of $4ca-b^2$

lab bhattacharjee
  • 274,582
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HINT:

$$\int\frac{ev+f}{av^2+bv+c}\space\text{d}v=$$ $$\int\left[\frac{e(2av+b)}{2a(av^2+bv+c)}-\frac{eb-2af}{2a(av^2+bv+c)}\right]\space\text{d}v=$$ $$\frac{e}{2a}\int\frac{2av+b}{av^2+bv+c}\space\text{d}v+\left(f-\frac{eb}{2a}\right)\int\frac{1}{av^2+bv+c}\space\text{d}v=$$

Substitute $u=av^2+bv+c$ and $\text{d}u=\left(2av+b\right)\space\text{d}v$:

$$\frac{e}{2a}\int\frac{1}{u}\space\text{d}u+\left(f-\frac{eb}{2a}\right)\int\frac{1}{av^2+bv+c}\space\text{d}v=$$ $$\frac{e\ln\left|u\right|}{2a}+\left(f-\frac{eb}{2a}\right)\int\frac{1}{av^2+bv+c}\space\text{d}v=$$ $$\frac{e\ln\left|av^2+bv+c\right|}{2a}+\left(f-\frac{eb}{2a}\right)\int\frac{1}{av^2+bv+c}\space\text{d}v$$

For the second integral, complete the square....

Jan Eerland
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