What is the integral of 1/(u^2+4) ?

The integral of 1/(u^2+4) is 1/2 arctan(u/2)+C

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integral of 1/(u^2+4)

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

\frac{1}{2} arctan (\frac{u}{2}) + C

Solution steps

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

Apply Integral Substitution

= \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

= \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)

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

Substitute back

v = \frac{u}{2}

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

Add a constant to the solution

= \frac{1}{2} arctan (\frac{u}{2}) + C

x \to 0 + lim ((5 x)^{x})

x \to \infty lim (\frac{1}{x} - 1)

\int - 6 tan^{2} (x) d x

x^{3} \frac{d y}{d x} + (2 - 3 x^{2}) y = x^{3}

x \to 1 lim (\frac{x ^{3}}{x - 1})

What is the integral of 1/(u^2+4) ?
The integral of 1/(u^2+4) is 1/2 arctan(u/2)+C