# Homework Help: Hard integral

1. Apr 22, 2005

### kennis2

i cant resolve this integral
2x^3-2x^2+1/x^2-x-2?

2. Apr 22, 2005

### Muzza

Is that

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

?

If so, use polynomial longdivision and then partial fractions to transform the expression into something which can be integrated more easily.

3. Apr 22, 2005

### kennis2

yeah, can you so kind to do some procedures.. cuz i did with partial fractions
but the result is not correct =(
thx much!

4. Apr 22, 2005

### Ba

Partial fractions don't work because the denominator is less than the numerator, use Muzza's idea and use long division.

5. Apr 22, 2005

### Hurkyl

Staff Emeritus
Long division is part of the partial fractions algorithm!

6. Apr 22, 2005

### p53ud0 dr34m5

$$\int \frac{2x^3-2x^2+1}{x^2-x-2}$$
$$\frac{2x^3-2x^2+1}{x^2-x-2}=2x+\frac{2x+1}{2x^3-2x^2+1}$$
$$\int 2xdx + \int \frac{2x+1}{2x^3-2x^2+1}dx$$
now, use partial fractions on the second half. it shouldnt be too hard!

7. Apr 22, 2005

### dextercioby

The denominator is $(x-2)(x+1)$,so it shouldn't be 2 difficult to get the simple fractions...

Daniel.

8. Apr 24, 2005

### Hippo

Did you mean

$$\frac{2x^3-2x^2+1}{x^2-x-2}=2x+\frac{4x+1}{x^2-x-2}$$?

9. Apr 25, 2005

### dextercioby

Yes,of course,yours is correct.He couldn't have changed the denominator.I mean he did,but it was wrong.

Daniel.

10. Jun 29, 2011

### xbef

This is correct.

Partial Fraction Decompisition for $\frac{4x+1}{x^2-x-2}$ is:

$\frac{4x+1}{(x+1)(x-2)}$ = $\frac{1}{x+1} + \frac{3}{x-2}$

New Integral is:
$\int (2x + \frac{1}{x+1} + \frac{3}{x-2})dx$

so,

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

11. Dec 22, 2013

### CalculusManiac

x^2 + 3 Log[2 - x] + Log[1 + x]