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Step-by-step Solution

Integral of $\frac{x^3-x^2+x-3}{x+1}$ with respect to x

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Answer

$\frac{x^{3}}{3}-x^2+3x-6\ln\left|x\right|-6\ln\left|x+1\right|+C_0$

Step-by-step explanation

Problem to solve:

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

Let's divide the polynomial by $x+1$ using synthetic division (also known as Ruffini's rule). First, write all the coefficients of the polynomial in the numerator in descending order based on grade (putting a zero if a term doesn't exist). Then, take the first coefficient ($1$) and multiply it by the root of the denominator ($-1$). Add the result to the second coefficient and multiply this by $-1$ and so on

$\left|\begin{matrix}1 & -1 & 1 & -3 \\ & -1 & 2 & -3 \\ 1 & -2 & 3 & -6\end{matrix}\right|-1$
2

In the last row appear the new coefficients of the polynomial. Use these coefficients to rewrite the new polynomial with a lower grade, and the remainder ($-6$) divided by the divisor

$\int\left(x^{2}-2x+3-6x^{-1}+\frac{-6}{x+1}\right)dx$

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Answer

$\frac{x^{3}}{3}-x^2+3x-6\ln\left|x\right|-6\ln\left|x+1\right|+C_0$