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

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 \to 0 - lim (\frac{2 ^{x} - 1}{x})

x \to 0 - lim (e^{(\frac{1}{x})})

in v erse l a pl a ce \frac{4}{4 s ^{2} + 1}

\int \frac{1}{4} e^{4 x} d x

\frac{d}{d x} (- \frac{1}{3} x^{- 3})

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