# Thread: stuck on this long division problem

1. ## stuck on this long division problem

hi, this is as far as I got:

how do I integrate the rest? Also is my long division right?

2. Your long division is o.k.

Hint:

Obtain the derivative of ln(x^2+4)...

3. $\int \frac{(4+x^3)}{(4+x^2)} dx
$

For the integrand $\frac{(x^3+4)}{(x^2+4)}$, do long division:

$= \int (x-(4 \frac{(x-1))}{(x^2+4)}) dx$

Integrate the sum term by term and factor out constants:

$= \int x dx-4 \int \frac{(x-1)}{(x^2+4)} dx
$

Expanding the integrand $\frac{(x-1)}{(x^2+4)}$ gives

$\frac{x}{(x^2+4)}-\frac{1}{(x^2+4)}$:

$= \int x dx-4 \int \frac{(x}{(x^2+4)}-\frac{1}{(x^2+4)} dx
$

Integrate the sum term by term and factor out constants:

$= 4 \int \frac{1}{(x^2+4)} dx-4 \int \frac{x}{(x^2+4)} dx+ \int x dx
$

For the integrand $\frac{x}{(x^2+4)}$}, substitute $u = x^2+4$ and $du = 2 x dx$:

$= -2 \int \frac{1}{u} du+4 \int \frac{1}{(x^2+4)} dx+ \int x dx
$

The integral of $\frac{1}{(x^2+4)}$ is $\frac{1}{2} arctan(\frac{x}{2})$:

$= -2 \int \frac{1}{u} du+2 arctan(\frac{x}{2})+ \int x dx
$

The integral of $\frac{1}{u}$ is $ln(u)$:

$= -2 ln(u)+2 arctan(\frac{x}{2})+ \int x dx
$

The integral of $x$ is $\frac{x^2}{2}$:

$= -2 ln(u)+\frac{x^2}{2}+2arctan(\frac{x}{2})+C
$

Substitute back for $u = x^2+4$:

$=\frac {x^2}{2}-2 ln(x^2+4)+2 arctan(\frac{x}{2})+C$

4. hi, i was doing good until

the integral of is :

how do you know this? If i did this problem, i would have ended up substituting u for (x^2+4)

5. Originally Posted by softballchick
hi, i was doing good until

the integral of is :

how do you know this? If i did this problem, i would have ended up substituting u for (x^2+4)
It's very famous integral...

You can sole it by a trigonometric substitution: x=tan(u)