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

The integral of x/(x^2+8) is 1/2 ln|x^2+8|+C

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integral of x/(x^2+8)

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

\frac{1}{2} ln x^{2} + 8 + C

Solution steps

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

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

Add a constant to the solution

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

8 x^{\frac{3}{2}} + 6 x^{\frac{1}{2}} + 4^{- \frac{1}{2}}

\frac{\partial}{\partial y} (3 x y^{2} e^{4 yz})

x \to 0 + lim (tan^{x} (8 x))

f (x) = x^{2}, (- 1, 1)

What is the integral of x/(x^2+8) ?
The integral of x/(x^2+8) is 1/2 ln|x^2+8|+C