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

The integral of 4/(x^2+9) is 4/3 arctan(x/3)+C

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integral of 4/(x^2+9)

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

\frac{4}{3} arctan (\frac{x}{3}) + C

Solution steps

\int \frac{4}{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{1}{x ^{2} + 9} d x

Apply Integral Substitution

= 4 \cdot \int \frac{1}{3 ( u ^{2} + 1 )} d u

Take the constant out:

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

= 4 \cdot \frac{1}{3} \cdot \int \frac{1}{u ^{2} + 1} d u

Use the common integral:

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

= 4 \cdot \frac{1}{3} arctan (u)

Substitute back

u = \frac{x}{3}

= 4 \cdot \frac{1}{3} arctan (\frac{x}{3})

Simplify

4 \cdot \frac{1}{3} arctan (\frac{x}{3}) : \frac{4}{3} arctan (\frac{x}{3})

= \frac{4}{3} arctan (\frac{x}{3})

Add a constant to the solution

= \frac{4}{3} arctan (\frac{x}{3}) + C

y^{^{''}} - 4 y^{^{'}} + 4 y = 6 t^{2} e^{2 t}

f (x) = ln (e x)

x \to - 3 lim (\frac{2 x + 8}{x + 3})

\frac{d}{d x} (\frac{x ^{4} - 12 x ^{2}}{( x ^{2} - 4 ) ^{2}})

\frac{\partial}{\partial x} (arcsin (\frac{a}{b} sin (x)))

What is the integral of 4/(x^2+9) ?
The integral of 4/(x^2+9) is 4/3 arctan(x/3)+C