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

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

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integral of 4/(x^2+16)

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

arctan (\frac{x}{4}) + C

Solution steps

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

Take the constant out:

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

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

Apply Integral Substitution

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

Take the constant out:

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

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

Use the common integral:

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

= 4 \cdot \frac{1}{4} arctan (u)

Substitute back

u = \frac{x}{4}

= 4 \cdot \frac{1}{4} arctan (\frac{x}{4})

Simplify

4 \cdot \frac{1}{4} arctan (\frac{x}{4}) : arctan (\frac{x}{4})

= arctan (\frac{x}{4})

Add a constant to the solution

= arctan (\frac{x}{4}) + C

\int \frac{2 ( x + 2 )}{5} d x

f (x) = \frac{x ^{3} + 2}{3}, at x = - 3

\frac{d}{d x} (x^{4} + y^{4})

\frac{d}{d x} (\frac{x}{x ^{3} + 1})

f (x) = cos^{2} (2 x)

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