What is the integral of 5/(16+x^2) ?

The integral of 5/(16+x^2) is 5/4 arctan(x/4)+C

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integral of 5/(16+x^2)

\int \frac{5}{16 + x ^{2}} d x

\frac{5}{4} arctan (\frac{x}{4}) + C

Solution steps

\int \frac{5}{16 + x ^{2}} d x

Take the constant out:

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

= 5 \cdot \int \frac{1}{16 + x ^{2}} d x

Apply Integral Substitution

= 5 \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

= 5 \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)

= 5 \cdot \frac{1}{4} arctan (u)

Substitute back

u = \frac{x}{4}

= 5 \cdot \frac{1}{4} arctan (\frac{x}{4})

Simplify

5 \cdot \frac{1}{4} arctan (\frac{x}{4}) : \frac{5}{4} arctan (\frac{x}{4})

= \frac{5}{4} arctan (\frac{x}{4})

Add a constant to the solution

= \frac{5}{4} arctan (\frac{x}{4}) + C

ln (ln (8 s))

f (x) = 4 x^{4} - x^{6}

y = (6 x^{3} + 3 x + 1)^{10}

\int_{0}^{1} (x + y) d x

\int \frac{e ^{t} e ^{t} + 2}{e ^{t} + 6} d t

What is the integral of 5/(16+x^2) ?
The integral of 5/(16+x^2) is 5/4 arctan(x/4)+C