What is the integral of (64)/(x^2+16) ?

The integral of (64)/(x^2+16) is 16arctan(x/4)+C

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integral of (64)/(x^2+16)

\int \frac{64}{x ^{2} + 16} d x

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

Solution steps

\int \frac{64}{x ^{2} + 16} d x

Take the constant out:

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

= 64 \cdot \int \frac{1}{x ^{2} + 16} d x

Apply Integral Substitution

= 64 \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

= 64 \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)

= 64 \cdot \frac{1}{4} arctan (u)

Substitute back

u = \frac{x}{4}

= 64 \cdot \frac{1}{4} arctan (\frac{x}{4})

Simplify

64 \cdot \frac{1}{4} arctan (\frac{x}{4}) : 16 arctan (\frac{x}{4})

= 16 arctan (\frac{x}{4})

Add a constant to the solution

= 16 arctan (\frac{x}{4}) + C

\int \frac{1}{1 - 64 t ^{2}} d t

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What is the integral of (64)/(x^2+16) ?
The integral of (64)/(x^2+16) is 16arctan(x/4)+C