I know the integration when in the reciprocal there's only degree $1$, but what about degree $2$? Take an example, $$\int\frac{x \, \mathrm{d}x}{a+bx^2}$$
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$\begingroup$ Are $a$ and $b$ constants? or functions of $x$? $\endgroup$– TravisJJul 3, 2015 at 14:39
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$\begingroup$ Yes, they are constants $\endgroup$– AneekJul 3, 2015 at 14:41
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3$\begingroup$ Do u-substitution with $u=a + bx^2$ $\endgroup$– DougJul 3, 2015 at 14:44
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$\begingroup$ Try a $u$ substitution with $u=a+bx^2$ then $du=2bx dx$ or $\frac{1}{2b}du = xdx$. $\endgroup$– TravisJJul 3, 2015 at 14:44
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$\begingroup$ In the future, refrain from MathJax only titles. $\endgroup$– Zain PatelJul 3, 2015 at 15:31
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3 Answers
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For this, put $v=a+bx^{2}$ then $dv = 2bxdx$ from which \begin{eqnarray} \int \frac{xdx}{a+bx^{2}} &=& \frac{1}{2b}\int \frac{1}{v}dv \\ &=& \frac{1}{2b}\ln | v |+ c \\ &=& \frac{1}{2b}\ln |a+bx^{2}|+c \\ \end{eqnarray}
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$\begingroup$ Sorry, I gave mistyped the question a bit, here's the actual one $\endgroup$– AneekJul 3, 2015 at 14:42
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1$\begingroup$ Quick comment. It should be $\ln|v|$ not $\ln (v)$. $\endgroup$– user223391Jul 3, 2015 at 15:20
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$\begingroup$ @avid19 Good point, I'll re-edit to reflect requirements. Thanks. $\endgroup$ Jul 6, 2015 at 8:03
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$$\int\frac{xdx}{a+bx^2}=\frac1{2b}\int\frac{2bxdx}{a+bx^2}=\frac1{2b}\int\frac{d(a+bx^2)}{a+bx^2}=\frac1{2b}\ln(a+bx^2)+C$$
answered Jul 3, 2015 at 14:51
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Let $u = a+bx^2$ so that $\mathrm{d}u =2bx \, \mathrm{d}x \implies {\color{blue}{x \, \mathrm{d}x = \frac{1}{2b} \, \mathrm{d}u}}$. Then our integral becomes $$\int \frac{\color{blue}{x \, \mathrm{d}x}}{a+bx^2} = \frac{1}{2b}\int \frac{1}{u} \, \mathrm{d}u$$
This evaluates to $$\frac{1}{2b} \ln |u| + \mathrm{C} = \frac{1}{2b} \ln |a+bx^2| + \mathrm{C}$$
answered Jul 3, 2015 at 15:35