What is the integral of 1/(-8sqrt(x)-8x) ?

The integral of 1/(-8sqrt(x)-8x) is -1/4 ln|sqrt(x)+1|+C

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integral of 1/(-8sqrt(x)-8x)

\int \frac{1}{- 8 x - 8 x} d x

- \frac{1}{4} ln x + 1 + C

Solution steps

\int \frac{1}{- 8 x - 8 x} d x

Apply u-substitution

: \int - \frac{1}{4 ( u + 1 )} d u

= \int - \frac{1}{4 ( u + 1 )} d u

Take the constant out:

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

= - \frac{1}{4} \cdot \int \frac{1}{u + 1} d u

Apply u-substitution

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

= - \frac{1}{4} \cdot \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= - \frac{1}{4} ln ∣ v ∣

Substitute back

= - \frac{1}{4} ln x + 1

Add a constant to the solution

= - \frac{1}{4} ln x + 1 + C

f (x) = x^{3} - 4 x, at x = - 2

\int_{2}^{4} \frac{2 x + 5}{5 - x} d x

y^{^{'}} = x e^{- y}

3 y^{^{''}} + 5 y^{^{'}} - 2 y = 0

x \to 8 - lim (ln (8 - x))

What is the integral of 1/(-8sqrt(x)-8x) ?
The integral of 1/(-8sqrt(x)-8x) is -1/4 ln|sqrt(x)+1|+C