What is the integral of (37x+5)/((6x+1)(x-1)) ?

The integral of (37x+5)/((6x+1)(x-1)) is 1/6 ln|6x+1|+6ln|x-1|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (37x+5)/((6x+1)(x-1))

\int \frac{37 x + 5}{( 6 x + 1 ) ( x - 1 )} d x

\frac{1}{6} ln ∣ 6 x + 1 ∣ + 6 ln ∣ x - 1 ∣ + C

Solution steps

\int \frac{37 x + 5}{( 6 x + 1 ) ( x - 1 )} d x

Take the partial fraction of

\frac{37 x + 5}{( 6 x + 1 ) ( x - 1 )} : \frac{1}{6 x + 1} + \frac{6}{x - 1}

= \int \frac{1}{6 x + 1} + \frac{6}{x - 1} 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}{6 x + 1} d x + \int \frac{6}{x - 1} d x

\int \frac{1}{6 x + 1} d x = \frac{1}{6} ln ∣ 6 x + 1 ∣

\int \frac{6}{x - 1} d x = 6 ln ∣ x - 1 ∣

= \frac{1}{6} ln ∣ 6 x + 1 ∣ + 6 ln ∣ x - 1 ∣

Add a constant to the solution

= \frac{1}{6} ln ∣ 6 x + 1 ∣ + 6 ln ∣ x - 1 ∣ + C

f (x) = x \cdot ln (x)

\frac{\partial}{\partial x} (x^{8} + 6 x y^{7})

6 + \frac{1}{x ^{2}}

y^{^{''}} + 4 y = 3 cos (2 t) - 5 sin (2 t)

y = x^{4} - 4 x^{2} + 10, at x = 1

What is the integral of (37x+5)/((6x+1)(x-1)) ?
The integral of (37x+5)/((6x+1)(x-1)) is 1/6 ln|6x+1|+6ln|x-1|+C