What is the integral of (4x)/(x^2+5) ?

The integral of (4x)/(x^2+5) is 2ln|x^2+5|+C

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integral of (4x)/(x^2+5)

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

2 ln x^{2} + 5 + C

Solution steps

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

= 4 \cdot \frac{1}{2} ln x^{2} + 5

Simplify

4 \cdot \frac{1}{2} ln x^{2} + 5 : 2 ln x^{2} + 5

= 2 ln x^{2} + 5

Add a constant to the solution

= 2 ln x^{2} + 5 + C

(x^{2} + x y - y^{2}) d x + x y d y = 0

x \to 0 lim ((x - 3) x)

sin^{5} (3 x^{2} - x + 13)

\frac{d}{d x} ((\frac{3}{4})^{x})

(5 x + 3) + (5 y - 5) y^{^{'}} = 0

What is the integral of (4x)/(x^2+5) ?
The integral of (4x)/(x^2+5) is 2ln|x^2+5|+C