# Integration leading to ln II

#### paulmdrdo

##### Active member
i tried to solve this and I got stuck somewhere in my solution. I need help.

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

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!

#### MarkFL

Staff member
Re: integration leading to ln II

Using long division on the integrand, you should be able to show that:

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

Now, integrate term by term.

#### paulmdrdo

##### Active member
Re: integration leading to ln II

Using long division on the integrand, you should be able to show that:

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

Now, integrate term by term.
did i use a wrong method in my solution? and if it isn't can you help me continue solving the problem using my method above.

#### MarkFL

Staff member
Re: integration leading to ln II

did i use a wrong method in my solution? and if it isn't can you help me continue solving the problem using my method above.
I would abandon what you first tried in lieu of this much easier method. But that's just me.

#### topsquark

##### Well-known member
MHB Math Helper
Re: integration leading to ln II

$\displaystyle \int\frac{u-1}{u}du-\frac{2}{3}\int\frac{x}{u}du+\frac{5}{3}\int\frac{du}{u}-\frac{2}{3}\int\frac{du}{u\,x^2}$
You are a bit scrambled in your use of the u substitution. Generally when a substitution is made we note the following:
$$\displaystyle u = f(x) \text{ --> } x = f^{-1}(u)$$
$$\displaystyle du = f'(x) dx \text{ --> } dx = f'^{-1}(u) du$$

The notation is a bit abstract so let me use your example:
$$\displaystyle u = x^3 + 1 \text{ --> } x = (u - 1)^{1/3}$$
$$\displaystyle du = 3x^2 dx \text{ --> } dx = \frac{1}{3}(u - 1)^{-2/3}du$$

Note that the x's are all on one side of the equation and the u's are all on the other. This means that you can now proceed from x integrals directly to u integrals.

Try looking at your problem again in light of these relations. I'm suspecting you made a substitution error.

-Dan