What is the integral of 1/(4x^2+16) ?

The integral of 1/(4x^2+16) is 1/8 arctan(x/2)+C

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integral of 1/(4x^2+16)

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

\frac{1}{8} arctan (\frac{x}{2}) + C

Solution steps

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

Apply Integral Substitution

= \int \frac{1}{8 ( u ^{2} + 1 )} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int \frac{1}{u ^{2} + 1} d u

Use the common integral:

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

= \frac{1}{8} arctan (u)

Substitute back

u = \frac{2}{4} x

= \frac{1}{8} arctan (\frac{2}{4} x)

Simplify

\frac{1}{8} arctan (\frac{2}{4} x) : \frac{1}{8} arctan (\frac{x}{2})

= \frac{1}{8} arctan (\frac{x}{2})

Add a constant to the solution

= \frac{1}{8} arctan (\frac{x}{2}) + C

\frac{d}{d t} (t cos (2 t))

4 x + \frac{1}{x}

x sin (x^{2}) y^{^{'}} + y (4 x^{2} cos (x^{2}) - sin (x^{2})) = 0

\int x^{2} + 25 d x

\int \frac{3}{9 - x ^{2}} d x

What is the integral of 1/(4x^2+16) ?
The integral of 1/(4x^2+16) is 1/8 arctan(x/2)+C