Solving the Integral 2x^3-2x^2+1/x^2-x-2

[kennis2]

i can't resolve this integral
2x^3-2x^2+1/x^2-x-2?

 

[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.

 

[kennis2]

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

 

[Ba]

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

 

[Hurkyl]

Long division is part of the partial fractions algorithm!

 

[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 shouldn't be too hard!

 

[dextercioby]

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



Daniel.

 

[Hippo]

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

Did you mean

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

 

[dextercioby]

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

Daniel.

 

[xbef]

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

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,

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

 

[CalculusManiac]

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