Question

integrate (4(x - 4))/(x ^ 2 - 2x - 3) dx​

Answer 1

Jawab:

[tex]\displaystyle 5\ln|x+1|-\ln|x-3|+C[/tex]

Penjelasan dengan langkah-langkah:

Factorize the denominator and solve by partial fraction deconposition.

[tex]\displaystyle \int \frac{4(x-4)}{x^2-2x-3}~dx=\int \frac{4x-16}{(x+1)(x-3)}~dx[/tex]

Distinct linear factors

[tex]\begin{aligned}\frac{4x-16}{(x+1)(x-3)}&\:=\frac{A}{x+1}+\frac{B}{x-3}\\4x-16\:&=A(x-3)+B(x+1)\\4x-16\:&=(A+B)x+(-3A+B)\end{aligned}[/tex]

Find A and B

[tex]\displaystyle\left\{\begin{matrix}A+B=4\\ -3A+B=-16\end{matrix}\right.[/tex]

-3A + B = -16

A + B = 4

_________

-4A = -20 → A = 5

Substitute to A + B = 4 then B = -1 we get

[tex]\displaystyle \frac{4x-16}{(x+1)(x-3)}=\frac{5}{x+1}-\frac{1}{x-3}[/tex]

Solve the problem

[tex]\begin{aligned}\int \frac{4(x-4)}{x^2-2x-3}~dx&\:=\int \frac{4x-16}{(x+1)(x-3)}~dx\\\:&=\int \left ( \frac{5}{x+1}-\frac{1}{x-3} \right )dx\\\:&=5\ln|x+1|-\ln|x-3|+C\end{aligned}[/tex]