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

• kennis2
In summary, the conversation is about resolving an integral and using polynomial long division and partial fractions to transform the expression into something that can be integrated more easily. The final result is x^2 + ln|x+1| + 3*ln|x-2| + Cx^2 + 3 Log[2 - x] + Log[1 + x].
kennis2
i can't resolve this integral
2x^3-2x^2+1/x^2-x-2?

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.

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

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

Long division is part of the partial fractions algorithm!

$$\int \frac{2x^3-2x^2+1}{x^2-x-2}$$
$$\frac{2x^3-2x^2+1}{x^2-x-2}=2x+\frac{2x+1}{2x^3-2x^2+1}$$
$$\int 2xdx + \int \frac{2x+1}{2x^3-2x^2+1}dx$$
now, use partial fractions on the second half. it shouldn't be too hard!

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

Daniel.

p53ud0 dr34m5 said:
$$\frac{2x^3-2x^2+1}{x^2-x-2}=2x+\frac{2x+1}{2x^3-2x^2+1}$$

Did you mean

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

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

Daniel.

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

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

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

## What is an integral?

An integral is a mathematical concept used to calculate the area under a curve or the accumulation of a quantity over a given interval. It is the inverse operation of differentiation and is denoted by the symbol ∫.

## How do I solve an integral?

To solve an integral, you need to find the antiderivative of the given function. This can be done using integration techniques such as substitution, integration by parts, or partial fractions. Once the antiderivative is found, you can then evaluate the integral using the limits of integration.

## What is the difference between definite and indefinite integrals?

A definite integral has specific limits of integration and gives a numerical value as the result. An indefinite integral does not have limits of integration and gives a function as the result. Definite integrals are used to calculate areas or accumulated quantities, while indefinite integrals are used to find a general function that satisfies a given derivative.

## What is the order of operations for solving an integral?

The order of operations for solving an integral is the same as for any mathematical expression: perform any operations inside parentheses, evaluate exponents, multiply and divide from left to right, and finally add and subtract from left to right.

## How can I check my solution to an integral?

You can check your solution to an integral by taking the derivative of your antiderivative and seeing if it matches the original function. If it does, then your solution is correct. You can also use online integral calculators or check your answer using a graphing calculator.

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