What is the integral of (5x)/(x^2+8) ?

The integral of (5x)/(x^2+8) is 5/2 ln|x^2+8|+C

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integral of (5x)/(x^2+8)

\int \frac{5 x}{x ^{2} + 8} d x

\frac{5}{2} ln x^{2} + 8 + C

Solution steps

\int \frac{5 x}{x ^{2} + 8} 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} + 8} 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} + 8

= 5 \cdot \frac{1}{2} ln x^{2} + 8

Simplify

5 \cdot \frac{1}{2} ln x^{2} + 8 : \frac{5}{2} ln x^{2} + 8

= \frac{5}{2} ln x^{2} + 8

Add a constant to the solution

= \frac{5}{2} ln x^{2} + 8 + C

x \to 4 lim (x + 5 - 3)

\frac{\partial}{\partial x} (e^{x - 2 y})

\int (\frac{1}{1 + cos ( x )}) d x

x \to 0 lim (x^{2} c x c x)

\frac{d}{d x} (tanh (x - 1))

What is the integral of (5x)/(x^2+8) ?
The integral of (5x)/(x^2+8) is 5/2 ln|x^2+8|+C