What is the integral from 2 to 5 of 1/(2-5t) ?

The integral from 2 to 5 of 1/(2-5t) is 1/5 (3ln(2)-ln(23))

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integral from 2 to 5 of 1/(2-5t)

\int_{2}^{5} \frac{1}{2 - 5 t} d t

\frac{1}{5} (3 ln (2) - ln (23))

Decimal

- 0.21121 \dots

Solution steps

\int_{2}^{5} \frac{1}{2 - 5 t} d t

Apply u-substitution

: \int_{- 8}^{- 23} - \frac{1}{5 u} d u

= \int_{- 8}^{- 23} - \frac{1}{5 u} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{- 23}^{- 8} - \frac{1}{5 u} d u

Take the constant out:

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

= - (- \frac{1}{5} \cdot \int_{- 23}^{- 8} \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - (- \frac{1}{5} [ln ∣ u ∣]_{- 23}^{- 8})

Simplify

= \frac{1}{5} [ln ∣ u ∣]_{- 23}^{- 8}

Compute the boundaries:

3 ln (2) - ln (23)

= \frac{1}{5} (3 ln (2) - ln (23))

9 y^{^{''}} - 6 y^{^{'}} + 4 y = 0

\frac{d}{d x} ((1 - x) e^{- x})

x \to 0 lim (\frac{4 sin ( 3 x )}{2 tan ( 5 x )})

x \to (\infty) lim (\frac{ln ( x )}{x})

x \to 1 lim (\frac{( x ^{2} )}{x - 1})

What is the integral from 2 to 5 of 1/(2-5t) ?
The integral from 2 to 5 of 1/(2-5t) is 1/5 (3ln(2)-ln(23))