What is the integral of (24)/((399x-1)(399+1)) ?

The integral of (24)/((399x-1)(399+1)) is 1/6650 ln|399x-1|+C

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integral of (24)/((399x-1)(399+1))

\int \frac{24}{( 399 x - 1 ) ( 399 + 1 )} d x

\frac{1}{6650} ln ∣ 399 x - 1 ∣ + C

Solution steps

\int \frac{24}{( 399 x - 1 ) ( 399 + 1 )} d x

Take the constant out:

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

= \frac{24}{399 + 1} \cdot \int \frac{1}{399 x - 1} d x

\frac{24}{399 + 1} = \frac{3}{50}

= \frac{3}{50} \cdot \int \frac{1}{399 x - 1} d x

Apply u-substitution

: \int \frac{1}{399 u} d u

= \frac{3}{50} \cdot \int \frac{1}{399 u} d u

Take the constant out:

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

= \frac{3}{50} \cdot \frac{1}{399} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{3}{50} \cdot \frac{1}{399} ln ∣ u ∣

Substitute back

u = 399 x - 1

= \frac{3}{50} \cdot \frac{1}{399} ln ∣ 399 x - 1 ∣

Simplify

\frac{3}{50} \cdot \frac{1}{399} ln ∣ 399 x - 1 ∣ : \frac{1}{6650} ln ∣ 399 x - 1 ∣

= \frac{1}{6650} ln ∣ 399 x - 1 ∣

Add a constant to the solution

= \frac{1}{6650} ln ∣ 399 x - 1 ∣ + C

arcsin (u)

\frac{d}{d x} (3 x^{2} + 2 x y - y^{2})

f (x) = (3 - x)^{4} (x + 5)^{3}

\frac{d y}{d t} = y \cdot ln (y)

\frac{\partial}{\partial x} (7 x e^{x^{2} + y^{2}})

What is the integral of (24)/((399x-1)(399+1)) ?
The integral of (24)/((399x-1)(399+1)) is 1/6650 ln|399x-1|+C