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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (6x)/(x^2+4)

\int \frac{6 x}{x ^{2} + 4} d x

3 ln x^{2} + 4 + C

Solution steps

\int \frac{6 x}{x ^{2} + 4} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 6 \cdot \int \frac{x}{x ^{2} + 4} d x

Apply u-substitution

: \int \frac{1}{2 u} d u

= 6 \cdot \int \frac{1}{2 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 6 \cdot \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 6 \cdot \frac{1}{2} ln ∣ u ∣

Substitute back

u = x^{2} + 4

= 6 \cdot \frac{1}{2} ln x^{2} + 4

Simplify

6 \cdot \frac{1}{2} ln x^{2} + 4 : 3 ln x^{2} + 4

= 3 ln x^{2} + 4

Add a constant to the solution

= 3 ln x^{2} + 4 + C

(- 2, 7), (0, 5)

\frac{d}{d v} (\frac{u}{v})

x \to 0 lim ((1 + 7 x)^{\frac{5}{x}})

\frac{\partial}{\partial y} (- x e^{- 3 x y})

\int_{0}^{1} (1 + x)^{3} d x

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