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\int\frac{1-3x}{x^2+2x-8}dx

Integral of (1-3x)/(2x-8+x^2)

Answer

$-\frac{13}{6}\ln\left|4+x\right|-\frac{5}{6}\ln\left|x-2\right|+C_0$

Step-by-step explanation

Problem

$\int\frac{1-3x}{x^2+2x-8}dx$
1

Factor the trinomial $-8+2x+x^2$ finding two numbers that multiply to form $-8$ and added form $2$

$\begin{matrix}\left(-2\right)\left(4\right)=-8\\ \left(-2\right)+\left(4\right)=2\end{matrix}$

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Answer

$-\frac{13}{6}\ln\left|4+x\right|-\frac{5}{6}\ln\left|x-2\right|+C_0$

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$\int\frac{1-3x}{x^2+2x-8}dx$

Main topic:

Integrals by partial fraction expansion

Used formulas:

6. See formulas

Time to solve it:

~ 1.0 seconds