What is the integral of x/(x^2+6x+8) ?

The integral of x/(x^2+6x+8) is 1/2 ln|x^2+6x+8|+3(1/2 ln|x+4|-1/2 ln|x+2|)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of x/(x^2+6x+8)

\int \frac{x}{x ^{2} + 6 x + 8} d x

\frac{1}{2} ln x^{2} + 6 x + 8 + 3 (\frac{1}{2} ln ∣ x + 4 ∣ - \frac{1}{2} ln ∣ x + 2 ∣) + C

Solution steps

\int \frac{x}{x ^{2} + 6 x + 8} d x

Complete the square

x^{2} + 6 x + 8 : (x + 3)^{2} - 1

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

Apply u-substitution

: \int \frac{u - 3}{u ^{2} - 1} d u

= \int \frac{u - 3}{u ^{2} - 1} d u

Expand

\frac{u - 3}{u ^{2} - 1} : \frac{u}{u ^{2} - 1} - \frac{3}{u ^{2} - 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{u}{u ^{2} - 1} d u - \int \frac{3}{u ^{2} - 1} d u

\int \frac{u}{u ^{2} - 1} d u = \frac{1}{2} ln u^{2} - 1

\int \frac{3}{u ^{2} - 1} d u = - 3 (\frac{1}{2} ln ∣ u + 1 ∣ - \frac{1}{2} ln ∣ u - 1 ∣)

= \frac{1}{2} ln u^{2} - 1 - (- 3 (\frac{1}{2} ln ∣ u + 1 ∣ - \frac{1}{2} ln ∣ u - 1 ∣))

Substitute back

u = x + 3

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

Simplify

\frac{1}{2} ln (x + 3)^{2} - 1 - (- 3 (\frac{1}{2} ln ∣ x + 3 + 1 ∣ - \frac{1}{2} ln ∣ x + 3 - 1 ∣)) : \frac{1}{2} ln x^{2} + 6 x + 8 + 3 (\frac{1}{2} ln ∣ x + 4 ∣ - \frac{1}{2} ln ∣ x + 2 ∣)

= \frac{1}{2} ln x^{2} + 6 x + 8 + 3 (\frac{1}{2} ln ∣ x + 4 ∣ - \frac{1}{2} ln ∣ x + 2 ∣)

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 6 x + 8 + 3 (\frac{1}{2} ln ∣ x + 4 ∣ - \frac{1}{2} ln ∣ x + 2 ∣) + C

\int \frac{1}{16 x ^{2} - 121} d x

\int sin^{7} (x) \cdot cos^{8} (x) d x

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

\int \frac{x ^{2} + x + 5}{x ^{3} + x ^{2} + 4 x + 4} d x

\int \frac{3 x ^{2}}{e ^{3 x}} d x

What is the integral of x/(x^2+6x+8) ?
The integral of x/(x^2+6x+8) is 1/2 ln|x^2+6x+8|+3(1/2 ln|x+4|-1/2 ln|x+2|)+C