please integrate for me the following function, it comes to me in the exam, and I put my answer on it, but I don't know if it's true or not
I did analyze the to
and I multiply it to
and the integration become's easier, then I did the rest.
I didn't complete it because my computer doesn't have a battery, and the kids coming by near the thing anytime and anytime will be off, so I submit it fast
On
$$\int\sqrt{x}\left(x+2\right)^2\space\text{d}x=$$
Substitute $u=\sqrt{x}$ and $\text{d}u=\frac{1}{2\sqrt{x}}\space\text{d}x$:
$$2\int u^2\left(u^2+2\right)^2\space\text{d}u=$$ $$2\int\left[u^6+4u^4+4u^2\right]\space\text{d}u=$$ $$2\left[\int u^6\space\text{d}u+4\int u^4\space\text{d}u+4\int u^2\space\text{d}u\right]=$$
Now, use the fact that:
$$\int s^b\space\text{d}s=\frac{s^{b+1}}{b+1}+\text{C}$$
$$2\left[\frac{u^7}{7}+\frac{4u^5}{5}+\frac{4u^3}{3}\right]+\text{C}=$$ $$2\left[\frac{\left(\sqrt{x}\right)^7}{7}+\frac{4\left(\sqrt{x}\right)^5}{5}+\frac{4\left(\sqrt{x}\right)^3}{3}\right]+\text{C}=$$ $$2\left[\frac{x^{\frac{7}{2}}}{7}+\frac{4x^{\frac{5}{2}}}{5}+\frac{4x^{\frac{3}{2}}}{3}\right]+\text{C}$$
it is $$\sqrt{x}(x^2+4x+4)=x^2\cdot x^{1/2}+4x\cdot x^{1/2}+4x^{1/2}=x^{5/2}+4x^{3/2}+4x^{1/2}$$ and now you can integrate this term