como fazer essa integral indefinida pelo metodo de frações parciais

tem um app que faz a integral so tirar a foto da integral ele resolve procure no google play

Para resolver a integral dada, realizaremos os cálculos abaixo:

\(\begin{align} & \int_{{}}^{{}}{f(x)=}\int_{{}}^{{}}{\frac{18}{{{(4{{x}^{2}}+9)}^{2}}}} \\ & \int_{{}}^{{}}{\frac{18}{{{(4{{x}^{2}}+9)}^{2}}}}=18\int_{{}}^{{}}{\frac{1}{{{(4{{x}^{2}}+9)}^{2}}}} \\ & x=\frac{3}{2}\tan u \\ & dx=\frac{3{{\sec }^{2}}u}{2}du \\ & \int_{{}}^{{}}{\frac{18}{{{(4{{x}^{2}}+9)}^{2}}}}=18\int_{{}}^{{}}{\frac{1}{{{(4{{x}^{2}}+9)}^{2}}}} \\ & \int_{{}}^{{}}{\frac{18}{{{(4{{x}^{2}}+9)}^{2}}}}=\int_{{}}^{{}}{\frac{1}{{{\left( 4{{\left( \frac{3}{2}\tan u \right)}^{2}}+9 \right)}^{2}}\left( \frac{3{{\sec }^{2}}u}{2}du \right)}} \\ & \int_{{}}^{{}}{\frac{18}{{{(4{{x}^{2}}+9)}^{2}}}}=\frac{1}{54}\int_{{}}^{{}}{\frac{1}{2}+\int_{{}}^{{}}{\frac{\cos 2u}{2}}} \\ & \int_{{}}^{{}}{\frac{18}{{{(4{{x}^{2}}+9)}^{2}}}}=\frac{1}{54}\left( \frac{u}{2}+\frac{1}{2}\int_{{}}^{{}}{\cos 2udu} \right) \\ & \int_{{}}^{{}}{\frac{18}{{{(4{{x}^{2}}+9)}^{2}}}}=\frac{{{\tan }^{-1}}\frac{2x}{3}}{108}+\frac{\frac{2\frac{2}{3}x}{\sqrt{1+\frac{4}{9}{{x}^{2}}}}\left( \frac{1}{\sqrt{1+\frac{4}{9}{{x}^{2}}}} \right)}{216} \\ & \int_{{}}^{{}}{\frac{18}{{{(4{{x}^{2}}+9)}^{2}}}}=18\left( \frac{{{\tan }^{-1}}\frac{2x}{3}}{108} \right)+\frac{x}{162\left( 1+\frac{4{{x}^{2}}}{9} \right)}+C \\ \end{align}\ \)