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

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

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integral of x/(9+x^2)

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

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

Solution steps

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

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

Add a constant to the solution

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

\frac{d}{d x} (\frac{6 cos ( x )}{1 - sin ( x )})

\int_{1}^{1} x^{2} d x

\int - e^{- x} cos (3 x) d x

\frac{d}{d x} (2 x^{3} e^{2 x})

\frac{\partial}{\partial y} (x^{2} e^{- yz})

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