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How do we integrate the following integral?

$$\int \frac{1}{x+\large\frac{1}{x^2}}\,dx\quad\text{where}\;\;x\ne-1$$

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The function is $\dfrac{x^2}{x^3+1}$. Let $u=x^3+1$.

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+1, bravo good hint. – Rustyn Mar 7 '13 at 9:07
Why give hints, if others will come along and do a complete solution in a matter of minutes? – GEdgar Mar 7 '13 at 20:16
@GEdgar: Well, this was much of the way to an answer. And as to why give hints, one does one's thing, others can do theirs. – André Nicolas Mar 7 '13 at 20:21

$$\quad\int \frac{1}{x+\large\frac{1}{x^2}}\,dx \quad = \quad\int \frac{1}{\Large\frac {x^3+ 1}{x^2}}\,dx\quad= \quad\int \frac {\color{blue}{\bf x^2\,dx}}{\color{red}{\bf x^3 + 1}}\quad(x \neq -1)$$

Let us substitute $\;\;\color{red}{\bf u} = \color{red}{\bf x^3 + 1} \;\implies\; du = 3x^2\,dx \;\implies \;\color{blue}{\bf\dfrac 13 du} = \color{blue}{\bf x^2\,dx}$.

Substituting equivalent expressions gives us:

$$\int \color{blue}{\bf \frac 13} \frac{\color{blue}{\bf du}}{\color{red}{\bf u}} \;\;= \;\;\frac 13 \ln|u| + C \;\;= \;\;\frac 13 \ln|x^3 + 1| + C, \;\;x\neq -1$$

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One question that I always have is - how do you determine the substitution variable? – bryansis2010 Mar 7 '13 at 9:18
After manipulating to get the integrand in the form $\dfrac{x^2}{x^3 + 1}$, I note that $\frac d{dy}(x^3 + 1)\,$ is $3x^2$, and so we're all set up for substituting $u = x^3 + 2$,(for the denominator), because $du$ is clearly $3x^2\,dx$, which virtually matches the numerator, by adjusting to let $x^2 = \frac 13\,du$. Then we have a nice, readily integrable integrand of the form $\dfrac 13 \int \dfrac {du}{u}$. That is, our original integrand is equivalent to $\dfrac 13 \int \frac{{d}{dx}(x^3 + 1)}{x^3 + 1} = 1/3\ln|x^3 + 1|+C$ – amWhy Mar 7 '13 at 17:44
@Rajesh Is this clear now? Feel free to accept an answer. (Just click on the $✓$ to the left of the answer)$\quad$ ;-) – amWhy Mar 8 '13 at 17:34

The integral is equivalent to

$$\int dx \frac{x^2}{1+x^3} = \frac{1}{3} \int \frac{d(x^3)}{1+x^3} = \frac{1}{3} \log{(1+x^3)} + C$$

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