What is the integral of (16)/(x^2+4) ?

The integral of (16)/(x^2+4) is 8arctan(x/2)+C

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integral of (16)/(x^2+4)

\int \frac{16}{x ^{2} + 4} d x

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

Solution steps

\int \frac{16}{x ^{2} + 4} d x

Take the constant out:

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

= 16 \cdot \int \frac{1}{x ^{2} + 4} d x

Apply Integral Substitution

= 16 \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

= 16 \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)

= 16 \cdot \frac{1}{2} arctan (u)

Substitute back

u = \frac{x}{2}

= 16 \cdot \frac{1}{2} arctan (\frac{x}{2})

Simplify

16 \cdot \frac{1}{2} arctan (\frac{x}{2}) : 8 arctan (\frac{x}{2})

= 8 arctan (\frac{x}{2})

Add a constant to the solution

= 8 arctan (\frac{x}{2}) + C

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What is the integral of (16)/(x^2+4) ?
The integral of (16)/(x^2+4) is 8arctan(x/2)+C