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

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

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

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

\frac{5}{2} ln x^{2} + 4 + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 5 \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

= 5 \cdot \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 5 \cdot \frac{1}{2} ln ∣ u ∣

Substitute back

u = x^{2} + 4

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

Simplify

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

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (5 x sin (x y))

\frac{\partial}{\partial x} (ln (8 y e^{x y}))

\int \frac{1}{( 25 - x ^{2} ) ^{2}} d x

3 x^{2} - x^{3}

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

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