What is the integral of 1/(16sqrt(x)+16x) ?

The integral of 1/(16sqrt(x)+16x) is 1/8 ln|sqrt(x)+1|+C

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integral of 1/(16sqrt(x)+16x)

\int \frac{1}{16 x + 16 x} d x

\frac{1}{8} ln x + 1 + C

Solution steps

\int \frac{1}{16 x + 16 x} d x

Apply u-substitution

: \int \frac{1}{8 ( u + 1 )} d u

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

Apply u-substitution

: \int \frac{1}{v} d v

= \frac{1}{8} \cdot \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{8} ln ∣ v ∣

Substitute back

= \frac{1}{8} ln x + 1

Add a constant to the solution

= \frac{1}{8} ln x + 1 + C

y = x x, at x = 16

x \to 3 lim (10 - x)

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What is the integral of 1/(16sqrt(x)+16x) ?
The integral of 1/(16sqrt(x)+16x) is 1/8 ln|sqrt(x)+1|+C