Evaluating $\int_0^2 \frac{x^2}{x^3 + 1} \,\mathrm{d}x$

Question

I don't understand how to get from the first to the second step here and get $1/3$ in front.

In the second step $g(x)$ substitutes $x^3 + 1$.

\begin{align*} \int_0^2 \frac{x^2}{x^3 + 1} \,\mathrm{d}x &= \frac{1}{3} \int_{0}^{2} \frac{1}{g(x)} g'(x) \,\mathrm{d}x = \frac{1}{3} \int_{1}^{9} \frac{1}{u} \,\mathrm{d}u \\ &= \left. \frac{1}{3} \ln(u) \,\right|_{1}^{9} = \frac{1}{3} ( \ln 9 - \ln 1 ) = \frac{\ln 9}{3} \end{align*}

Answer 1

The derivative of $g(x)$ is $g'(x)=3x^2$. The constant $3$ is not present in the first expression but:

$\frac{1}{3}g'(x)=\frac{1}{3}\cdot 3x^2=x^2$

Answer 2

Let $u:=x^3+1$. Then it follows that

$$\frac{\operatorname{d}u}{\operatorname{d}x}=3x^2 \Longleftrightarrow \operatorname{d}x=\frac{\operatorname{d}u}{3x^2}.$$

Putting this into the integral leads to

$$\int_{0}^2 \frac{x^2}{x^3+1}\operatorname{d}x= \int_{1}^9 \frac{x^2}{u} \cdot \frac{\operatorname{d}u}{3x^2}= \frac{1}{3} \int_{1}^9 \frac{\operatorname{d}u}{u}$$

and then the rest like above.

Answer 3

Looking at the denominator, you define $g(x)=x^3+1$. This means that $g'(x)=3x^2$.

Your goal is to manipulate the integrand so that it is in the form $g'(x)/g(x)$, that is, $3x^2/(x^3+1)$. As it stands, the factor $3$ you need is missing. You can't just throw it in, because that changes the integrand. But you can throw it in and pull it out at the same time, since that amounts to multiplying by $1$, which doesn't change anything: $$\frac{x^2}{x^3-1}=\frac{\overbrace{(\frac13\cdot 3)}^1 x^2}{x^3-1}$$ $$=\frac{\frac13\cdot (3 x^2)}{x^3-1}$$ $$=\tfrac13\cdot \frac{3x^2}{x^3-1}$$ So you proceed from here.

Generally, if you need a missing constant factor of $c$ in the integrand, you can put it in if you also compensate by putting its reciprocal $\frac1c$ in as well. And remember, constant factors can be pulled to the outside of the integral. The rule is

$$\int f(x)\; dx = \frac1c\int cf(x)\; dx$$