# How do I evaluate the integral int7x^2 ln(x) dx?

Jan 26, 2015

I'd use integration by parts: since $\left(f g\right) ' = f ' g + f g '$, by integrating this relation one has that $f g = \setminus \int f ' g + \setminus \int f g '$.

When you have to integrate the product of two functions, integration by parts is often a good strategy: if one of the two factors is easily integrable/derivable, you can try and see if the integral becomes easier. In your case, I'd use the fact that
$\left(1\right) \setminus \int f ' g = f g - \setminus \int f g '$

So, we have the following:
$f \left(x\right) = \frac{7}{3} {x}^{3} , \setminus \setminus \setminus f ' \left(x\right) = 7 {x}^{2}$
$g \left(x\right) = \setminus \ln \left(x\right) , \setminus \setminus \setminus g ' \left(x\right) = \frac{1}{x}$.

And using the equation labeled as $\left(1\right)$, you have that

$\setminus \int 7 {x}^{2} \setminus \ln \left(x\right) = \frac{7}{3} {x}^{3} \setminus \ln \left(x\right) - \setminus \int \frac{7}{3} {x}^{3} \frac{1}{x}$

The andantage is evident: the integral of $\frac{7}{3} {x}^{3} \frac{1}{x} = \frac{7}{3} {x}^{2}$ is obviously $\frac{7}{9} {x}^{3}$, and so our solution is

$\setminus \int 7 {x}^{2} \setminus \ln \left(x\right) = \frac{7}{3} {x}^{3} \setminus \ln \left(x\right) - \frac{7}{9} {x}^{3} + c$