Answer
$$\int^{1}_{0}\frac{x^3}{\sqrt{x^4+9}}dx=\frac{1}{2}(\sqrt{10}-3)$$
$$\int^{0}_{-1}\frac{x^3}{\sqrt{x^4+9}}dx=-\frac{1}{2}(\sqrt{10}-3)$$
Work Step by Step
Part a) $$A=\int^{1}_{0}\frac{x^3}{\sqrt{x^4+9}}dx$$
Set $u=\sqrt{x^4+9}$, we then have $$du=\frac{(x^4+9)'}{2\sqrt{x^4+9}}dx=\frac{4x^3dx}{2\sqrt{x^4+9}}=\frac{2x^3dx}{\sqrt{x^4+9}}$$ $$\frac{x^3dx}{\sqrt{x^4+9}}=\frac{1}{2}du$$
For $x=0$, we have $u=3$ and for $x=1$, we have $u=\sqrt{10}$.
Therefore, $$A=\frac{1}{2}\int^{\sqrt10}_3du=\frac{1}{2}u\Big]^{\sqrt10}_3$$ $$A=\frac{1}{2}(\sqrt{10}-3)$$
Part b) $$A=\int_{-1}^{0}\frac{x^3}{\sqrt{x^4+9}}dx$$
We carry out a similar substitution as in part a).
For $x=0$, we have $u=3$ and for $x=-1$, we have $u=\sqrt{10}$.
Therefore, $$A=\frac{1}{2}\int_{\sqrt10}^3du=-\frac{1}{2}\int^{\sqrt{10}}_3du=-\frac{1}{2}(\sqrt{10}-3)$$
