What is the integral of (3x^2+8x-12)/(x^3+7x^2+12x) ?

The integral of (3x^2+8x-12)/(x^3+7x^2+12x) is -ln|x|+3ln|x+3|+ln|x+4|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (3x^2+8x-12)/(x^3+7x^2+12x)

\int \frac{3 x ^{2} + 8 x - 12}{x ^{3} + 7 x ^{2} + 12 x} d x

- ln ∣ x ∣ + 3 ln ∣ x + 3 ∣ + ln ∣ x + 4 ∣ + C

Solution steps

\int \frac{3 x ^{2} + 8 x - 12}{x ^{3} + 7 x ^{2} + 12 x} d x

Take the partial fraction of

\frac{3 x ^{2} + 8 x - 12}{x ^{3} + 7 x ^{2} + 12 x} : - \frac{1}{x} + \frac{3}{x + 3} + \frac{1}{x + 4}

= \int - \frac{1}{x} + \frac{3}{x + 3} + \frac{1}{x + 4} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int \frac{1}{x} d x + \int \frac{3}{x + 3} d x + \int \frac{1}{x + 4} d x

\int \frac{1}{x} d x = ln ∣ x ∣

\int \frac{3}{x + 3} d x = 3 ln ∣ x + 3 ∣

\int \frac{1}{x + 4} d x = ln ∣ x + 4 ∣

= - ln ∣ x ∣ + 3 ln ∣ x + 3 ∣ + ln ∣ x + 4 ∣

Add a constant to the solution

= - ln ∣ x ∣ + 3 ln ∣ x + 3 ∣ + ln ∣ x + 4 ∣ + C

\int x^{3} x^{4} + 8 d x

\int x e^{\frac{x ^{2}}{5}} d x

\int \frac{2 t ^{3} - 4 t ^{2} + 1}{2 t} d t

\int \frac{x ^{2} + x}{x - 1} d x

\int x^{3} sin (4 x^{4} - π) d x

What is the integral of (3x^2+8x-12)/(x^3+7x^2+12x) ?
The integral of (3x^2+8x-12)/(x^3+7x^2+12x) is -ln|x|+3ln|x+3|+ln|x+4|+C