Answer
$$\int^{1/2}_{0}x^3(1+9x^4)^{-3/2}dx=\frac{1}{90}$$
Work Step by Step
$$A=\int^{1/2}_{0}x^3(1+9x^4)^{-3/2}dx$$
We set $a=1+9x^4$, which means $$da=36x^3dx$$ $$x^3dx=\frac{1}{36}da$$
For $x=0$, $a=1$ and for $x=1/2$, $a=1+\frac{9}{2^4}=1+\frac{9}{16}=\frac{25}{16}$
Therefore,
$$A=\frac{1}{36}\int^{25/16}_1a^{-3/2}da$$ $$A=\Big(\frac{1}{36}\times\frac{a^{-1/2}}{-\frac{1}{2}}\Big)\Big]^{25/16}_1=-\frac{1}{18\sqrt a}\Big]^{25/16}_1$$ $$A=-\frac{1}{18}\Big(\frac{1}{\sqrt{\frac{25}{16}}}-\frac{1}{\sqrt1}\Big)$$ $$A=-\frac{1}{18}\Big(\frac{1}{\frac{5}{4}}-1\Big)=-\frac{1}{18}\Big(\frac{4}{5}-1\Big)$$ $$A=-\frac{1}{18}\Big(-\frac{1}{5}\Big)$$ $$A=\frac{1}{90}$$
