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

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

Popular >

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

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

2 ln x^{2} + 4 + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 4 \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

= 4 \cdot \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 4 \cdot \frac{1}{2} ln ∣ u ∣

Substitute back

u = x^{2} + 4

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

Simplify

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

= 2 ln x^{2} + 4

Add a constant to the solution

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

\frac{\partial}{\partial y} (2 x - 3 y + 5)

(x^{2} + 2) (x^{3} + 6)

\int cos (x) e^{- s x} d x

\frac{\partial}{\partial x} (x^{2} + 1 + z^{3})

(\frac{5}{8}, - 2), (- \frac{1}{4}, - \frac{1}{4})

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