Answer to Question #266487 in Calculus for Faru

Question

Integrate (3-x^2)/(x^2+3)^2

Answer 1

$\begin{aligned} &\int \frac{3-x^{2}}{\left(x^{2}+3\right)^{2}} \mathrm{~d} x \\ &=-\int \frac{x^{2}-3}{\left(x^{2}+3\right)^{2}} \mathrm{~d} x \\ &=-\int\left(\frac{x^{2}+3}{\left(x^{2}+3\right)^{2}}-\frac{6}{\left(x^{2}+3\right)^{2}}\right) \mathrm{d} x \\ &=-\int\left(\frac{1}{x^{2}+3}-\frac{6}{\left(x^{2}+3\right)^{2}}\right) \mathrm{d} x \\ &=-\int \frac{1}{x^{2}+3} \mathrm{~d} x-6 \int \frac{1}{\left(x^{2}+3\right)^{2}} \mathrm{~d} x \\ &=-\frac{\arctan \left(\frac{x}{\sqrt{3}}\right)}{\sqrt{3}}+6\left[\frac{x}{6\left(x^{2}+3\right)}+\frac{1}{6} \int \frac{1}{x^{2}+3} \mathrm{~d} x\right] \\ &[\because \frac{1}{\left(\mathrm{a} x^{2}+\mathrm{b}\right)^{\mathrm{n}}} \mathrm{d} x=\frac{2 \mathrm{n}-3}{2 \mathrm{~b}(\mathrm{n}-1)} \int \frac{1}{\left(\mathrm{a} x^{2}+\mathrm{b}\right)^{\mathrm{n}-1}} \mathrm{~d} x+\frac{x}{2 \mathrm{~b}(\mathrm{n}-1)\left(\mathrm{a} x^{2}+\mathrm{b}\right)^{\mathrm{n}-1}} ]\\ &=-\frac{\arctan \left(\frac{x}{\sqrt{3}}\right)}{\sqrt{3}}+ \frac{x}{x^{2}+3}+\frac{\arctan \left(\frac{x}{\sqrt{3}}\right)}{\sqrt{3}} =\frac{x}{x^{2}+3} \end{aligned}$