What is the integral of (4x)/(x^2+9) ?

The integral of (4x)/(x^2+9) is 2ln|x^2+9|+C

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integral of (4x)/(x^2+9)

\int \frac{4 x}{x ^{2} + 9} d x

2 ln x^{2} + 9 + C

Solution steps

\int \frac{4 x}{x ^{2} + 9} d x

Take the constant out:

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

= 4 \cdot \int \frac{x}{x ^{2} + 9} d x

Apply u-substitution

: \int \frac{1}{2 u} d u

= 4 \cdot \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

= 4 \cdot \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 4 \cdot \frac{1}{2} ln ∣ u ∣

Substitute back

u = x^{2} + 9

= 4 \cdot \frac{1}{2} ln x^{2} + 9

Simplify

4 \cdot \frac{1}{2} ln x^{2} + 9 : 2 ln x^{2} + 9

= 2 ln x^{2} + 9

Add a constant to the solution

= 2 ln x^{2} + 9 + C

y = 2 x sin (x) + x^{2} e^{x}

\frac{d}{d x} (- 2 x^{2} - 5 x + 2)

\int \frac{1}{x ^{2} - 81} d x

y^{^{'}} = - \frac{3 y}{t} - 2 - t^{- 4}, y (1) = 4

\frac{\partial}{\partial y} (\frac{x - 3}{y ^{2} + 1})

What is the integral of (4x)/(x^2+9) ?
The integral of (4x)/(x^2+9) is 2ln|x^2+9|+C