What is the integral of (x+5)/((x+3)(x-1)) ?

The integral of (x+5)/((x+3)(x-1)) is ln|x+3|+6(1/4 ln|x-1|-1/4 ln|x+3|)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x+5)/((x+3)(x-1))

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

ln ∣ x + 3 ∣ + 6 (\frac{1}{4} ln ∣ x - 1 ∣ - \frac{1}{4} ln ∣ x + 3 ∣) + C

Solution steps

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

Apply u-substitution

: \int \frac{u + 6}{u ( u + 4 )} d u

= \int \frac{u + 6}{u ( u + 4 )} d u

Expand

\frac{u + 6}{u ( u + 4 )} : \frac{u}{u ( u + 4 )} + \frac{6}{u ( u + 4 )}

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{u}{u ( u + 4 )} d u + \int \frac{6}{u ( u + 4 )} d u

\int \frac{u}{u ( u + 4 )} d u = ln ∣ u + 4 ∣

\int \frac{6}{u ( u + 4 )} d u = 6 (\frac{1}{4} ln ∣ u ∣ - \frac{1}{4} ln ∣ u + 4 ∣)

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

Substitute back

u = x - 1

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

Simplify

ln ∣ x - 1 + 4 ∣ + 6 (\frac{1}{4} ln ∣ x - 1 ∣ - \frac{1}{4} ln ∣ x - 1 + 4 ∣) : ln ∣ x + 3 ∣ + 6 (\frac{1}{4} ln ∣ x - 1 ∣ - \frac{1}{4} ln ∣ x + 3 ∣)

= ln ∣ x + 3 ∣ + 6 (\frac{1}{4} ln ∣ x - 1 ∣ - \frac{1}{4} ln ∣ x + 3 ∣)

Add a constant to the solution

= ln ∣ x + 3 ∣ + 6 (\frac{1}{4} ln ∣ x - 1 ∣ - \frac{1}{4} ln ∣ x + 3 ∣) + C

\frac{d}{d x} (15 ln (\frac{x}{80}))

\frac{\partial}{\partial y} (y^{x z} + 2 (yz)^{x})

y = - \frac{4 x}{x ^{2} + 1}

\frac{arccos ( x )}{x + 1}

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

What is the integral of (x+5)/((x+3)(x-1)) ?
The integral of (x+5)/((x+3)(x-1)) is ln|x+3|+6(1/4 ln|x-1|-1/4 ln|x+3|)+C