What is the integral of (10x)/(x^2-2x-24) ?

The integral of (10x)/(x^2-2x-24) is 10(2/5 ln|x+4|+3/5 ln|x-6|)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (10x)/(x^2-2x-24)

\int \frac{10 x}{x ^{2} - 2 x - 24} d x

10 (\frac{2}{5} ln ∣ x + 4 ∣ + \frac{3}{5} ln ∣ x - 6 ∣) + C

Solution steps

\int \frac{10 x}{x ^{2} - 2 x - 24} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 10 \cdot \int \frac{x}{x ^{2} - 2 x - 24} d x

Take the partial fraction of

\frac{x}{x ^{2} - 2 x - 24} : \frac{2}{5 ( x + 4 )} + \frac{3}{5 ( x - 6 )}

= 10 \cdot \int \frac{2}{5 ( x + 4 )} + \frac{3}{5 ( x - 6 )} 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

= 10 (\int \frac{2}{5 ( x + 4 )} d x + \int \frac{3}{5 ( x - 6 )} d x)

\int \frac{2}{5 ( x + 4 )} d x = \frac{2}{5} ln ∣ x + 4 ∣

\int \frac{3}{5 ( x - 6 )} d x = \frac{3}{5} ln ∣ x - 6 ∣

= 10 (\frac{2}{5} ln ∣ x + 4 ∣ + \frac{3}{5} ln ∣ x - 6 ∣)

Add a constant to the solution

= 10 (\frac{2}{5} ln ∣ x + 4 ∣ + \frac{3}{5} ln ∣ x - 6 ∣) + C

f (x) = \frac{e ^{x}}{x}

8 t

\frac{d ^{2}}{d x ^{2}} (x + 7)

\int x^{2} (x^{3} + 1)^{2} d x

g (x) = \frac{5}{6 x ^{3}}

What is the integral of (10x)/(x^2-2x-24) ?
The integral of (10x)/(x^2-2x-24) is 10(2/5 ln|x+4|+3/5 ln|x-6|)+C