What is the integral of 6/(x^2-7x+10) ?

The integral of 6/(x^2-7x+10) is 6(-1/3 ln|x-2|+1/3 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 6/(x^2-7x+10)

\int \frac{6}{x ^{2} - 7 x + 10} d x

6 (- \frac{1}{3} ln ∣ x - 2 ∣ + \frac{1}{3} ln ∣ x - 5 ∣) + C

Solution steps

\int \frac{6}{x ^{2} - 7 x + 10} d x

Take the constant out:

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

= 6 \cdot \int \frac{1}{x ^{2} - 7 x + 10} d x

Take the partial fraction of

\frac{1}{x ^{2} - 7 x + 10} : - \frac{1}{3 ( x - 2 )} + \frac{1}{3 ( x - 5 )}

= 6 \cdot \int - \frac{1}{3 ( x - 2 )} + \frac{1}{3 ( 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

= 6 (- \int \frac{1}{3 ( x - 2 )} d x + \int \frac{1}{3 ( x - 5 )} d x)

\int \frac{1}{3 ( x - 2 )} d x = \frac{1}{3} ln ∣ x - 2 ∣

\int \frac{1}{3 ( x - 5 )} d x = \frac{1}{3} ln ∣ x - 5 ∣

= 6 (- \frac{1}{3} ln ∣ x - 2 ∣ + \frac{1}{3} ln ∣ x - 5 ∣)

Add a constant to the solution

= 6 (- \frac{1}{3} ln ∣ x - 2 ∣ + \frac{1}{3} ln ∣ x - 5 ∣) + C

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

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

\int \frac{6 x}{e ^{x}} d x

\int 6 x + 5 cos (x) d x

\int cosh (4 x + 3) d x

What is the integral of 6/(x^2-7x+10) ?
The integral of 6/(x^2-7x+10) is 6(-1/3 ln|x-2|+1/3 ln|x-5|)+C