Lena Bell

2022-07-01

How is the integral calculator finding this ${\int }_{0}^{a}\frac{{a}^{2}-{x}^{2}}{\left({a}^{2}+{x}^{2}{\right)}^{2}}dx$ while I get an undefined expression?

conveneau71

Expert

Alternatively, both cloudy's suggestion and the following method (the third step uses $\theta =\mathrm{arcsinh}\frac{x}{a}$ and integration by parts) also avoid having to divide by 0:
$\begin{array}{rl}& \int \frac{{a}^{2}-{x}^{2}}{\left({a}^{2}+{x}^{2}{\right)}^{2}}\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x\\ =& \int \frac{2{a}^{2}}{\left({a}^{2}+{x}^{2}{\right)}^{2}}-\frac{1}{{a}^{2}+{x}^{2}}\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x\\ =& \frac{2}{a}\int {\mathrm{sech}}^{3}\left(\mathrm{arcsinh}\frac{x}{a}\right)\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}\left(\mathrm{arcsinh}\frac{x}{a}\right)-\frac{1}{a}\mathrm{arctan}\frac{x}{a}\\ =& \frac{1}{a}\left(\mathrm{sech}\theta \mathrm{tanh}\theta +\int \mathrm{sech}\theta \phantom{\rule{thinmathspace}{0ex}}\mathrm{d}\theta \right)-\frac{1}{a}\mathrm{arctan}\frac{x}{a}\\ =& \frac{1}{a}\left(\mathrm{sinh}\theta {\mathrm{sech}}^{2}\theta +\mathrm{arctan}\frac{x}{a}\right)-\frac{1}{a}\mathrm{arctan}\frac{x}{a}+C\\ =& \frac{1}{a}\left(\frac{x}{a}×\frac{1}{1+{\left(\frac{x}{a}\right)}^{2}}\right)+C\\ =& \frac{x}{{a}^{2}+{x}^{2}}+C.\end{array}$
Alternatively, use $\theta =\mathrm{arctan}\frac{x}{a}$ or integrate by reduction formula.

Do you have a similar question?