What is the integral of x/(x^2+9) ?

The integral of x/(x^2+9) is 1/2 ln|x^2+9|+C

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integral of x/(x^2+9)

\int \frac{x}{x ^{2} + 9} d x

\frac{1}{2} ln x^{2} + 9 + C

Solution steps

\int \frac{x}{x ^{2} + 9} 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} + 9

= \frac{1}{2} ln x^{2} + 9

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 9 + C

y^{^{'''}} + 2 y^{^{''}} - 10 y^{^{'}} - 20 y = 0

f (x) = x^{2} + x - 7

x \to 2 lim (x^{2} - 3 x + 4)

\frac{d}{d x} (\frac{x ^{6}}{5 - x ^{5}})

x = 9 y^{2}, x = 20 + 4 y^{2}

What is the integral of x/(x^2+9) ?
The integral of x/(x^2+9) is 1/2 ln|x^2+9|+C