Evaluate improper integrals

Question

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

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

2. $ \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. $$\begin{eqnarray} \int_{0}^1 \ln x \cdot x^\frac{-1}{3} &=& \ln x\cdot\frac32x^\frac23-\int_0^1\frac32x^\frac{-1}{3}\cr&=&\left[\frac32 x^\frac23 \cdot\left(\ln x - \frac32 \right)\right]_0^1\\&=&-\frac94-(???) \end{eqnarray} $$

Answer 1

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.

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Just a remark on vocabulary: primitive is the same as antiderivative or indefinite integral, look for example on Wikipedia.

Answer 2

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ 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}