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#### paulmdrdo

##### Active member

- May 13, 2013

- 386

$\displaystyle \int\frac{3x^5-2x^3+5x^2-2}{x^3+1}dx$

splitting the integrand i get,

$\displaystyle \int\frac{3x^5}{x^3+1}dx-\int\frac{2x^3}{x^3+1}dx+\int\frac{5x^2}{x^3+1}dx-\int\frac{2}{x^3+1}dx$

by letting

$\displaystyle u\,=\,x^3+1$

$\displaystyle du\,=\,3x^2dx$

$\displaystyle dx\,=\,\frac{du}{3x^2}$

$\displaystyle x^3\,=\,u-1$

by substituting i now have,

$\displaystyle \int\frac{3x^5}{u}\frac{du}{3x^2}-\int\frac{2x^3}{u}\frac{du}{3x^2}+\int\frac{5x^2}{u}\frac{du}{3x^2}-\int\frac{2}{u}\frac{du}{3x^2}$

then,

$\displaystyle \int\frac{x^3}{u}du-\frac{2}{3}\int\frac{x}{u}du+\frac{5}{3}\int\frac{du}{u}-\frac{2}{3}\int\frac{du}{u\,x^2}$

since x^3 = u-1, i rewrote the first integral as $\displaystyle \int\frac{u-1}{u}du$

now i have written all the integrals in terms of u except 2nd and 4th integral

$\displaystyle \int\frac{u-1}{u}du-\frac{2}{3}\int\frac{x}{u}du+\frac{5}{3}\int\frac{du}{u}-\frac{2}{3}\int\frac{du}{u\,x^2}$

getting the indefinite integral of 1st and 3rd integral i get,

$\displaystyle \int udu-\int\frac{du}{u}+\frac{5}{3}\int\frac{du}{u} = u-\ln|u|+\frac{5}{3}\ln|u|$

and substituting the value of u i now have,

$\displaystyle x^3-\ln|x^3+1|+\frac{5}{3}\ln|x^3+1|\,=\,x^3+\frac{2}{3}\ln|x^3+1|$

until here i couldn't continue. i don't know what to do with the 2nd and 4th integral.

can you pinpoint where I'm wrong. thanks!