What is the integral of (2000)/(5x+10) ?

The integral of (2000)/(5x+10) is 400ln|5x+10|+C

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integral of (2000)/(5x+10)

\int \frac{2000}{5 x + 10} d x

400 ln ∣ 5 x + 10 ∣ + C

Solution steps

\int \frac{2000}{5 x + 10} d x

Take the constant out:

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

= 2000 \cdot \int \frac{1}{5 x + 10} d x

Apply u-substitution

: \int \frac{1}{5 u} d u

= 2000 \cdot \int \frac{1}{5 u} d u

Take the constant out:

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

= 2000 \cdot \frac{1}{5} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2000 \cdot \frac{1}{5} ln ∣ u ∣

Substitute back

u = 5 x + 10

= 2000 \cdot \frac{1}{5} ln ∣ 5 x + 10 ∣

Simplify

2000 \cdot \frac{1}{5} ln ∣ 5 x + 10 ∣ : 400 ln ∣ 5 x + 10 ∣

= 400 ln ∣ 5 x + 10 ∣

Add a constant to the solution

= 400 ln ∣ 5 x + 10 ∣ + C

\int_{0}^{2} x 5 - x^{2} d x

\frac{( s + 2 )}{( s ^{2} + s + 1 )}

\frac{- 25}{s ^{2} + 6 s + 25}

f (x) = x^{2} - 8, at x = 4

x y^{^{'}} = 3 y - 6 x^{2}

What is the integral of (2000)/(5x+10) ?
The integral of (2000)/(5x+10) is 400ln|5x+10|+C