What is the integral of x/(x^2+6) ?

The integral of x/(x^2+6) is 1/2 ln|x^2+6|+C

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integral of x/(x^2+6)

\int \frac{x}{x ^{2} + 6} d x

\frac{1}{2} ln x^{2} + 6 + C

Solution steps

\int \frac{x}{x ^{2} + 6} 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} + 6

= \frac{1}{2} ln x^{2} + 6

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 6 + C

y^{^{''}} - 9 y^{^{'}} + 18 y = 0, y (0) = - 5, y^{^{'}} (0) = 2

f (x) = 9 x^{2} - x^{3}, (2, 28)

\frac{d}{d x} (\frac{1}{3 x ^{5}})

f (x) = (\frac{sin ( x )}{x})^{2}

\int 4 x^{\frac{2}{3}} d x

What is the integral of x/(x^2+6) ?
The integral of x/(x^2+6) is 1/2 ln|x^2+6|+C