1. ## integration

would you possibly be able to show me step by step how to integrate this with respect to x:

((-x)^2)/(1+x^3)

Thanks

2. Originally Posted by daaavo
would you possibly be able to show me step by step how to integrate this with respect to x:

((-x)^2)/(1+x^3)

Thanks
$\displaystyle \int \frac{(-x)^2}{1+x^3}dx=\int \frac{x^2}{1+x^3}dx$

let $\displaystyle u=1+x^3 \mbox{ then } du=3x^2dx \iff \frac{1}{3}du=x^2dx$

so

$\displaystyle \int \frac{x^2}{1+x^3}dx=\frac{1}{3} \int \frac{1}{u}du =\frac{1}{3}\ln|u|+C =\frac{1}{3}\ln|1+x^3|+C$

3. Originally Posted by daaavo
would you possibly be able to show me step by step how to integrate this with respect to x:

((-x)^2)/(1+x^3)

Thanks

$\displaystyle \int \frac{(-x)^2}{1+x^3} \, dx$

$\displaystyle \int \frac{x^2}{1+x^3} \, dx$

$\displaystyle \frac13 \int \frac{3x^2}{1+x^3} \, dx$

Now make the substitution $\displaystyle u = x^3 \Rightarrow 3x^2 \, dx = du$

$\displaystyle \frac13 \int \frac{1}{1+u} \, du = \frac13\ln |u + 1| + C= \frac13\ln|1+x^3| +C$