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

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

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

\frac{1}{2} ln x^{2} + 4 + C

Solution steps

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

Apply u-substitution

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

= \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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} ln ∣ u ∣

Substitute back

u = x^{2} + 4

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

Add a constant to the solution

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

5 x (x^{3} - 5 x + 5)

\frac{\partial}{\partial x} (4 ln (\frac{2 x y}{7 z}))

h (w) = \frac{4 w ^{5} - w}{w}

\int x e^{1 - x} d x

f (x) = x sin (\frac{1}{x})

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