What is the integral of 6/((4x-2)(x+1)) ?

The integral of 6/((4x-2)(x+1)) is ln|4x-2|-ln|4x+4|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 6/((4x-2)(x+1))

\int \frac{6}{( 4 x - 2 ) ( x + 1 )} d x

ln ∣ 4 x - 2 ∣ - ln ∣ 4 x + 4 ∣ + C

Solution steps

\int \frac{6}{( 4 x - 2 ) ( x + 1 )} 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}{( 4 x - 2 ) ( x + 1 )} d x

Apply u-substitution

: \int \frac{1}{u ( u + 6 )} d u

= 6 \cdot \int \frac{1}{u ( u + 6 )} d u

Take the partial fraction of

\frac{1}{u ( u + 6 )} : \frac{1}{6 u} - \frac{1}{6 ( u + 6 )}

= 6 \cdot \int \frac{1}{6 u} - \frac{1}{6 ( u + 6 )} d u

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}{6 u} d u - \int \frac{1}{6 ( u + 6 )} d u)

\int \frac{1}{6 u} d u = \frac{1}{6} ln ∣ u ∣

\int \frac{1}{6 ( u + 6 )} d u = \frac{1}{6} ln ∣ u + 6 ∣

= 6 (\frac{1}{6} ln ∣ u ∣ - \frac{1}{6} ln ∣ u + 6 ∣)

Substitute back

u = 4 x - 2

= 6 (\frac{1}{6} ln ∣ 4 x - 2 ∣ - \frac{1}{6} ln ∣ 4 x - 2 + 6 ∣)

Simplify

6 (\frac{1}{6} ln ∣ 4 x - 2 ∣ - \frac{1}{6} ln ∣ 4 x - 2 + 6 ∣) : ln ∣ 4 x - 2 ∣ - ln ∣ 4 x + 4 ∣

= ln ∣ 4 x - 2 ∣ - ln ∣ 4 x + 4 ∣

Add a constant to the solution

= ln ∣ 4 x - 2 ∣ - ln ∣ 4 x + 4 ∣ + C

\int 16 t e^{t} d t

\int (x^{3} - 4 x^{2} + 5) sin (4 x) d x

\int e^{x (a - s)} d x

\int \frac{5}{x ^{2}} d x

\int \frac{1}{4 x ^{2} - 16} d x

What is the integral of 6/((4x-2)(x+1)) ?
The integral of 6/((4x-2)(x+1)) is ln|4x-2|-ln|4x+4|+C