Question

integrate (3x ^ 3 - 2x ^ 2 - 16x + 20)/((x - 2)(x + 2)) dx​

Answer 1

Jawab:

[tex]\displaystyle \frac{3x^2}{2}-2x-5\ln|x+2|+\ln|x-2|+C[/tex]

Penjelasan dengan langkah-langkah:

Because degrees of polynomial numerator greater than denominator, this problem solved by long division. (x - 2)(x + 2) = x² - 4

[tex]\begin{array}{ccccc} & 3x & -2\\\cline{2-5} x^2-4/ & 3x^3 & -2x^2 & -16x & +20\\ & 3x^3 & & -12x\\\cline{2-5} & & -2x^2 & -4x & +20\\ & & -2x^2 & & +8\\\cline{3-5} & & & -4x & +12\end{array}[/tex]

Definition of polynomial

f(x) = (divisor)(quotient) + remainder

In this case f(x) is dividend, so we can write

[tex]\displaystyle \rm{\frac{dividend}{divisor}=quotient+\frac{remainder}{divisor}}[/tex]

therefore

[tex]\displaystyle \frac{3x^3-2x^2-16x+20}{x^2-4}=3x-2+\frac{-4x+12}{x^2-4}[/tex]

Now perform partial fraction decomposition for [tex]\displaystyle \frac{-4x+12}{x^2-4}[/tex]

[tex]\begin{aligned}\frac{-4x+12}{(x+2)(x-2)}&\:=\frac{A}{x+2}+\frac{B}{x-2}\\-4x+12\:&=A(x-2)+B(x+2)\\-4x+12\:&=(A+B)x+(-2A+2B)\end{aligned}[/tex]

Find A and B

-A + B = 6

A + B = -4

_______

-2A = 10 → A = -5

then

A + B = -4

-5 + B = -4 → B = 1

So

[tex]\displaystyle \frac{-4x+12}{(x+2)(x-2)}=-\frac{5}{x+2}+\frac{1}{x-2}[/tex]

and we get

[tex]\displaystyle \frac{3x^3-2x^2-16x+20}{x^2-4}=3x-2-\frac{5}{x+2}+\frac{1}{x-2}[/tex]

Solve this problem

[tex]\begin{aligned}\int \frac{3x^3-2x^2-16x+20}{(x-2)(x+2)}&\:=\int \left ( 3x-2-\frac{5}{x+2}+\frac{1}{x-2} \right )dx\\\:&=\frac{3}{2}x^2-2x-5\ln|x+2|+\ln|x-2|+C\end{aligned}[/tex]