Evaluate improper integrals I can't get an answer for these two questions... help?


*

*$ \int_{-2}^0 \frac{x^2}{\left(1+x^3\right) ^2} dx $

*$ \int_{0}^1 \frac{\ln x}{x^\frac13} dx$
For question 1 I tried integrating but it gets more and more complicated. I got until $ \int \frac{1}{6(1+x^3} \frac{6x^2(1+x^3}{(1+x^3)^2}$ I'm not sure what to do afterwards or if it is even correct.
For question 2 I got until here. $$\begin{eqnarray} \int_{0}^1 \ln x \cdot x^\frac{-1}{3} &=& \ln x\cdot\frac32x^\frac23-\int_0^1\frac32x^\frac{-1}{3}\cr&=&\left[\frac32 x^\frac23 \cdot\left(\ln x - \frac32 \right)\right]_0^1\\&=&-\frac94-(???) \end{eqnarray} $$
 A: Question 1:
Notice $(1+x^3)' = 3x^2$, thus you are integrating "almost" $\frac{u'}{u^2}$, whose primitive is $-\frac{1}{u}$.
However, $1+x^3=0$ for $x=-1$, and your function $x\to\frac{x^2}{(1+x^3)^2}$ is not integrable on $[-2,0]$. It has a primitive, easy to compute with the hint above, but not defined at $x=-1$, and you can't integrate on an interval containing $-1$.
Question 2:
Try change of variable $x=y^3$, thus $\mathrm{d}x=3y^2\mathrm{d}y$ and
$$ \int \frac{\ln x}{x^\frac13} \mathrm{d}x = \int \frac{\log (y^3)}{y}3y^2\mathrm{d}y=
9\int y\log (y)\mathrm{d}y$$
And you can easily integrate this by parts.
The function $x\to\frac{\log x}{x^{1/3}}$ is like above not defined for $x=0$, but unlike the function in question 1, this function is integrable at $x=0$, and continuous on $]0,1]$, thus the integral exists, and the hint above to compute its primitive will do the job.

Just a remark on vocabulary: primitive is the same as antiderivative or indefinite integral, look for example on Wikipedia.
A: $\newcommand{\+}{^{\dagger}}%
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The first one is trivial.
$\large\tt Second\ one$:
With $\mu > -1$:
\begin{align}
\color{#00f}{\large\int_{0}^{1}{\ln\pars{x} \over x^{1/3}}\,\dd x}&=
\lim_{\mu \to -1/3}\partiald{}{\mu}\int_{0}^{1}x^{\mu}\,\dd x
=\lim_{\mu \to -1/3}\partiald{}{\mu}\pars{1 \over \mu + 1}
=\lim_{\mu \to -1/3}\bracks{-\,{1 \over \pars{\mu + 1}^{2}}}
\\[3mm]&=-\,{1 \over \pars{-1/3 + 1}^{2}} = \color{#00f}{\large -\,{9 \over 4}}
\end{align}
