What is the integral of 5/(x^2+4) ?

The integral of 5/(x^2+4) is 5/2 arctan(x/2)+C

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integral of 5/(x^2+4)

\int \frac{5}{x ^{2} + 4} d x

\frac{5}{2} arctan (\frac{x}{2}) + C

Solution steps

\int \frac{5}{x ^{2} + 4} d x

Take the constant out:

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

= 5 \cdot \int \frac{1}{x ^{2} + 4} d x

Apply Integral Substitution

= 5 \cdot \int \frac{1}{2 ( u ^{2} + 1 )} d u

Take the constant out:

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

= 5 \cdot \frac{1}{2} \cdot \int \frac{1}{u ^{2} + 1} d u

Use the common integral:

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

= 5 \cdot \frac{1}{2} arctan (u)

Substitute back

u = \frac{x}{2}

= 5 \cdot \frac{1}{2} arctan (\frac{x}{2})

Simplify

5 \cdot \frac{1}{2} arctan (\frac{x}{2}) : \frac{5}{2} arctan (\frac{x}{2})

= \frac{5}{2} arctan (\frac{x}{2})

Add a constant to the solution

= \frac{5}{2} arctan (\frac{x}{2}) + C

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

x \to \infty lim (\frac{3 ^{x - 3}}{2 ^{n + 3}})

x \to 2 - lim (x + 4)

3 y (x^{2} - 1) d x + (x^{3} + 8 y - 3 x) d y = 0

y = x^{2} - 5

What is the integral of 5/(x^2+4) ?
The integral of 5/(x^2+4) is 5/2 arctan(x/2)+C