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

\frac{d}{d x} (18 x e^{x})

\frac{\partial}{\partial x} (\frac{x + 4}{y + 3})

144 (\frac{216}{x ^{2}} + \frac{x}{x + 24})

y^{^{'}} = e^{x} y^{7} + 2 y

\int \frac{ln ( x + 1 )}{x + 1} d x

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