1. ## Integrals

I just want to know if my final answer is correct since I don't have the solution for it.

$\displaystyle \int\frac{\sqrt{x} - 2x^2}{x} dx$

I obtained: $\displaystyle \frac{2x^\frac{3}{2}}{3}lnx - \frac{2x^3}{3} + C$

Also, I need some help figuring out this integral:

$\displaystyle \int x(x^2 + 1)^5 dx$ It is continuous from 0 to 1.

2. Originally Posted by TGS
I just want to know if my final answer is correct since I don't have the solution for it.

$\displaystyle \int\frac{\sqrt{x} - 2x^2}{x} dx$

I obtained: $\displaystyle \frac{2x^\frac{3}{2}}{3}lnx - \frac{2x^3}{3} + C$

Also, I need some help figuring out this integral:

$\displaystyle \int x(x^2 + 1)^5 dx$ It is continuous from 0 to 1.
No, the first is not correct.

Divide everything by $\displaystyle x$ to turn the integral into

$\displaystyle \int{x^{-\frac{1}{2}} - 2x\,dx}$

$\displaystyle = 2x^{\frac{1}{2}} - x^2 + C$

$\displaystyle = 2\sqrt{x} - x^2 + C$.

For the second, you need to use a $\displaystyle u$ substitution.

Let $\displaystyle u = x^2 + 1$ so that $\displaystyle \frac{du}{dx} = 2x$.

$\displaystyle \int{x(x^2 + 1)^5\,dx} = \frac{1}{2}\int{2x(x^2 + 1)^5\,dx}$

$\displaystyle = \frac{1}{2}\int{u^5\,\frac{du}{dx}\,dx}$

$\displaystyle = \frac{1}{2}\int{u^5\,du}$

$\displaystyle = \frac{1}{12}u^6 + C$

$\displaystyle = \frac{1}{2}(x^2 + 1)^6 + C$.

Alternatively, you could expand everything, but the expansion of $\displaystyle (x^2 + 1)^5$ is difficult unless you know the Binomial Theorem.