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

The integral of (3x+4)/((x+1)(x+2)) is 2ln|x+2|+ln|x+1|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (3x+4)/((x+1)(x+2))

\int \frac{3 x + 4}{( x + 1 ) ( x + 2 )} d x

2 ln ∣ x + 2 ∣ + ln ∣ x + 1 ∣ + C

Solution steps

\int \frac{3 x + 4}{( x + 1 ) ( x + 2 )} d x

Apply u-substitution

= \int \frac{3 u + 1}{u ( u + 1 )} d u

Expand

\frac{3 u + 1}{u ( u + 1 )} : \frac{3 u}{u ( u + 1 )} + \frac{1}{u ( u + 1 )}

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{3 u}{u ( u + 1 )} d u + \int \frac{1}{u ( u + 1 )} d u

\int \frac{3 u}{u ( u + 1 )} d u = 3 ln ∣ u + 1 ∣

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

= 3 ln ∣ u + 1 ∣ + ln ∣ u ∣ - ln ∣ u + 1 ∣

Substitute back

u = x + 1

= 3 ln ∣ x + 1 + 1 ∣ + ln ∣ x + 1 ∣ - ln ∣ x + 1 + 1 ∣

Simplify

= 2 ln ∣ x + 2 ∣ + ln ∣ x + 1 ∣

Add a constant to the solution

= 2 ln ∣ x + 2 ∣ + ln ∣ x + 1 ∣ + C

\int (3 x - 5)^{15} d x

\int \frac{21}{x 1 + 2 x} d x

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

\int 3 x sec^{2} (3 x) d x

\int \frac{2}{x ( x ^{2} + 6 ) ^{2}} d x

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