What is the integral of 3/(u^2+4) ?

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

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integral of 3/(u^2+4)

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

\frac{3}{2} arctan (\frac{u}{2}) + C

Solution steps

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

Take the constant out:

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

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

Apply Integral Substitution

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

Take the constant out:

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

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

Use the common integral:

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

= 3 \cdot \frac{1}{2} arctan (v)

Substitute back

v = \frac{u}{2}

= 3 \cdot \frac{1}{2} arctan (\frac{u}{2})

Simplify

3 \cdot \frac{1}{2} arctan (\frac{u}{2}) : \frac{3}{2} arctan (\frac{u}{2})

= \frac{3}{2} arctan (\frac{u}{2})

Add a constant to the solution

= \frac{3}{2} arctan (\frac{u}{2}) + C

x \to - 1 lim (\frac{e ^{x}}{1 + x})

\int (x^{2} - 1) x d x

\frac{\partial}{\partial x} (- 1 - 2 x)

(e^{- t})^{^{'}}

- 10 e^{- x^{2}} x

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