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

The integral of x/((x^2+4)) is 1/2 ln|x^2+4|+C

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

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

\frac{1}{2} ln x^{2} + 4 + C

Solution steps

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

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

Add a constant to the solution

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

\int 4 + 3 x d x

\frac{\partial}{\partial x} (x + sin (y + z))

5 + 2 x - 7 x^{3}

\frac{d}{d x} (\frac{1}{2 x ^{2}} + \frac{x ^{4}}{16})

\frac{d}{d x} ((x^{2} - x + \frac{3}{4} x^{4})^{2})

What is the integral of x/((x^2+4)) ?
The integral of x/((x^2+4)) is 1/2 ln|x^2+4|+C