What is the integral of x/(1+9x^2) ?

The integral of x/(1+9x^2) is 1/18 ln|1+9x^2|+C

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integral of x/(1+9x^2)

\int \frac{x}{1 + 9 x ^{2}} d x

\frac{1}{18} ln 1 + 9 x^{2} + C

Solution steps

\int \frac{x}{1 + 9 x ^{2}} d x

Apply u-substitution

: \int \frac{1}{18 u} d u

= \int \frac{1}{18 u} d u

Take the constant out:

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

= \frac{1}{18} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{18} ln ∣ u ∣

Substitute back

u = 1 + 9 x^{2}

= \frac{1}{18} ln 1 + 9 x^{2}

Add a constant to the solution

= \frac{1}{18} ln 1 + 9 x^{2} + C

\frac{e ^{- 2 s}}{s - 9}

\int cos^{3} (3 x) sin^{3} (3 x) d x

f (x) = (e^{x} + x^{π}) (x^{2} + x + 1)

y^{^{'}} = 0.1 y (10 - y)

\frac{\partial}{\partial x} (ln (\frac{a}{x}))

What is the integral of x/(1+9x^2) ?
The integral of x/(1+9x^2) is 1/18 ln|1+9x^2|+C