What is the integral of (x-3)/((x+4)(x-5)) ?

The integral of (x-3)/((x+4)(x-5)) is 7/9 ln|x+4|+2/9 ln|x-5|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x-3)/((x+4)(x-5))

\int \frac{x - 3}{( x + 4 ) ( x - 5 )} d x

\frac{7}{9} ln ∣ x + 4 ∣ + \frac{2}{9} ln ∣ x - 5 ∣ + C

Solution steps

\int \frac{x - 3}{( x + 4 ) ( x - 5 )} d x

Take the partial fraction of

\frac{x - 3}{( x + 4 ) ( x - 5 )} : \frac{7}{9 ( x + 4 )} + \frac{2}{9 ( x - 5 )}

= \int \frac{7}{9 ( x + 4 )} + \frac{2}{9 ( x - 5 )} 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{7}{9 ( x + 4 )} d x + \int \frac{2}{9 ( x - 5 )} d x

\int \frac{7}{9 ( x + 4 )} d x = \frac{7}{9} ln ∣ x + 4 ∣

\int \frac{2}{9 ( x - 5 )} d x = \frac{2}{9} ln ∣ x - 5 ∣

= \frac{7}{9} ln ∣ x + 4 ∣ + \frac{2}{9} ln ∣ x - 5 ∣

Add a constant to the solution

= \frac{7}{9} ln ∣ x + 4 ∣ + \frac{2}{9} ln ∣ x - 5 ∣ + C

\int sec (θ) (sec (θ) + tan (x)) d θ

\int \frac{1}{49 + 16 x ^{2}} d x

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

\int [x] d x

\int \frac{x ^{5}}{( 9 - x ^{6} ) ^{2}} d x

What is the integral of (x-3)/((x+4)(x-5)) ?
The integral of (x-3)/((x+4)(x-5)) is 7/9 ln|x+4|+2/9 ln|x-5|+C