Answer
The solution is $$\int_{0}^{5}\frac{x^2}{(5+2x)^2}dx=\frac{5}{3}-\frac{5}{4}\ln3.$$
Work Step by Step
To solve the integral
$$\int_{0}^{5}\frac{x^2}{(5+2x)^2}dx$$
we will use substitution $t=5+2x$ and then we will have $x=\frac{t-5}{2}, dx=dt/2$ and for the bounds of integration we have $t=5,t=15$. Putting all this into the integral we have:
$$\int_{0}^{5}\frac{x^2}{(5+2x)^2}dx=\int_{5}^{15}\frac{\left(\frac{t-5}{2}\right)^2}{t^2}\frac{dt}{2}=\int_{5}^{15}\frac{(t-5)^2}{8t^2}dt=\int_{5}^{15}\frac{t^2-10t+25}{8t^2}dt=\frac{1}{8}\int_{5}^{15}dt-\frac{5}{4}\int_{5}^{15}\frac{dt}{t}+\frac{25}{8}\int_{5}^{15}\frac{1}{t^2}dt.$$
From the table of integrals we have:
Integral 1: $$\frac{1}{8}\int_{5}^{15}dt=\frac{1}{8}\left.t\right|_{5}^{15}=\frac{1}{8}(15-5)=\frac{5}{4};$$
Integral 2:$$\frac{5}{4}\int_{5}^{15}\frac{dt}{t}=\frac{5}{4}\left.\ln|t|\right|_{5}^{15}=\frac{5}{4}(\ln15-\ln5)=\frac{5}{4}(\ln\frac{15}{5})=\frac{5}{4}\ln3;$$
Integral 3: $$\frac{25}{8}\int_{5}^{15}\frac{dt}{t^2}=-\frac{25}{8}\left.\frac{1}{t}\right|_{5}^{15}=-\frac{25}{8}\left(\frac{1}{15}-\frac{1}{5}\right)=\frac{5}{12}.$$
Finally, our original integral is:
$$\int_{0}^{5}\frac{x^2}{(5+2x)^2}dx=\frac{5}{4}-\frac{5}{4}\ln3+\frac{5}{12}=\frac{5}{3}-\frac{5}{4}\ln3.$$
