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))

integral

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

\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

x^{8} \cdot ln (x)

\frac{x}{( 1 + x ) ^{2}}

\frac{1}{e ^{- x}}

- tan^{2} (x)

\frac{1}{sec ( x )}

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