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

\frac{d}{d x} (\frac{1 - cos ( 2 x )}{1 + cos ( 2 x )})

\frac{d}{d x} (\frac{x ^{2} - 4 x + 3}{( x - 2 ) ^{2}})

f (t) = 13 e^{4 t} \cdot e^{2}

\frac{d}{d x} (4 e^{- 6 x^{2}})

\int \frac{5 ( 1 - tan ^{2} ( x ) )}{sec ^{2} ( x )} d x

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