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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{1}{4 u} d u

= \int \frac{1}{4 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{4} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{4} ln ∣ u ∣

Substitute back

u = 2 x^{2} + 4

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

Add a constant to the solution

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

\int \frac{csc ^{2} ( x )}{2 x} d x

\frac{d}{d x} (2 sin (x - 2))

\frac{d}{d x} (9 - x^{\frac{2}{3}})

(cos^{3} (x))^{^{'}}

f (x) = \frac{2}{x}, at x = - 4

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