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

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

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integral of 4/(16+x^2)

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

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

Solution steps

\int \frac{4}{16 + x ^{2}} 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}{16 + x ^{2}} 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 sin^{4} (v) cos^{3} (v) d v

x y^{^{'}} + y = 2 x y^{2}, y (1) = - 5

\frac{\partial}{\partial x} (e^{y} sin (x y))

\frac{d}{d x} (\frac{12 x ^{2} - x ^{4}}{( 4 - x ^{2} ) ^{2}})

f (x) = (x - 3) (2 x + 4)

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