What is the integral of 1/(-20sqrt(x)-20x) ?

The integral of 1/(-20sqrt(x)-20x) is -1/10 ln|sqrt(x)+1|+C

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integral of 1/(-20sqrt(x)-20x)

\int \frac{1}{- 20 x - 20 x} d x

- \frac{1}{10} ln x + 1 + C

Solution steps

\int \frac{1}{- 20 x - 20 x} d x

Apply u-substitution

: \int - \frac{1}{10 ( u + 1 )} d u

= \int - \frac{1}{10 ( u + 1 )} d u

Take the constant out:

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

= - \frac{1}{10} \cdot \int \frac{1}{u + 1} d u

Apply u-substitution

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

= - \frac{1}{10} \cdot \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= - \frac{1}{10} ln ∣ v ∣

Substitute back

= - \frac{1}{10} ln x + 1

Add a constant to the solution

= - \frac{1}{10} ln x + 1 + C

x y^{^{'}} + (1 + 2 x^{2}) y = x^{3} e^{- x^{2}}

\frac{d}{d x} (sec (7 x))

\frac{d}{d x} (2 x (x + 1)^{2})

x \to 3 + lim (\frac{x}{3 x - x ^{3}})

\frac{d}{d x} ((sin (x))^{2})

What is the integral of 1/(-20sqrt(x)-20x) ?
The integral of 1/(-20sqrt(x)-20x) is -1/10 ln|sqrt(x)+1|+C