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

The integral of 1/(u^2+9) is 1/3 arctan(u/3)+C

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integral of 1/(u^2+9)

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

\frac{1}{3} arctan (\frac{u}{3}) + C

Solution steps

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

Apply Integral Substitution

= \int \frac{1}{3 ( v ^{2} + 1 )} d v

Take the constant out:

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

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

Use the common integral:

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

= \frac{1}{3} arctan (v)

Substitute back

v = \frac{u}{3}

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

Add a constant to the solution

= \frac{1}{3} arctan (\frac{u}{3}) + C

\frac{d}{d x} (lo g_{e} (sin (x)))

t (t cos (t) - sin (t))

n = 1 \sum \infty \frac{- 1 ^{n}}{n}

x \to 2 lim (\frac{x ^{2} - 3 x + 2}{x ^{2} + 5 x - 14})

t^{2} \cdot e^{- 4 t}

What is the integral of 1/(u^2+9) ?
The integral of 1/(u^2+9) is 1/3 arctan(u/3)+C