What is the integral of 1/(u^2+3) ?

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

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integral of 1/(u^2+3)

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

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

Solution steps

\int \frac{1}{u ^{2} + 3} 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 x}{d y} = \frac{y ^{3} - 3 x}{y}

n = 1 \sum \infty \frac{4}{2 n}

\frac{d ^{3}}{d x ^{3}} (\frac{6}{x ^{2}})

\int (y^{2} cos (x) - 3 x^{2} y - 2 x) d x

\frac{\partial}{\partial z} (ln (x y))

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