integration

**daaavo** · April 16th 2008, 10:10 AM

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

**TheEmptySet** · April 16th 2008, 10:16 AM

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

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

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

so

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

**Isomorphism** · April 16th 2008, 10:20 AM

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

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

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

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

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

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

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