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

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

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

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

- \frac{1}{2} arctan (\frac{x}{4}) + C

Solution steps

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

Take the constant out:

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

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

Apply Integral Substitution

= - 2 \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

= - 2 \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)

= - 2 \cdot \frac{1}{4} arctan (u)

Substitute back

u = \frac{x}{4}

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

Simplify

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

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

Add a constant to the solution

= - \frac{1}{2} arctan (\frac{x}{4}) + C

\frac{x}{( 1 + 4 x ^{2} ) ^{2}}

\frac{\partial}{\partial z} (x^{2} y + z)

y^{^{''}} + 10 y^{^{'}} + 9 y = 0, y (0) = 1, y^{^{'}} (0) = 0

(x sin (x))^{^{'}}

\frac{9}{s ^{2} + 9 s + 9}

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