What is the integral of 1/(x^2+3x-4) ?

The integral of 1/(x^2+3x-4) is 1/5 ln|x-1|-1/5 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 1/(x^2+3x-4)

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

\frac{1}{5} ln ∣ x - 1 ∣ - \frac{1}{5} ln ∣ x + 4 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{1}{x ^{2} + 3 x - 4} : \frac{1}{5 ( x - 1 )} - \frac{1}{5 ( x + 4 )}

= \int \frac{1}{5 ( x - 1 )} - \frac{1}{5 ( 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}{5 ( x - 1 )} d x - \int \frac{1}{5 ( x + 4 )} d x

\int \frac{1}{5 ( x - 1 )} d x = \frac{1}{5} ln ∣ x - 1 ∣

\int \frac{1}{5 ( x + 4 )} d x = \frac{1}{5} ln ∣ x + 4 ∣

= \frac{1}{5} ln ∣ x - 1 ∣ - \frac{1}{5} ln ∣ x + 4 ∣

Add a constant to the solution

= \frac{1}{5} ln ∣ x - 1 ∣ - \frac{1}{5} ln ∣ x + 4 ∣ + C

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

x \to 0 lim (4 - x)

\frac{\partial}{\partial x} (ln (x y) + ln (yz) + 5 ln (x z))

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

x^{2} - 4, x + 2

What is the integral of 1/(x^2+3x-4) ?
The integral of 1/(x^2+3x-4) is 1/5 ln|x-1|-1/5 ln|x+4|+C