What is the integral from 1 to 2 of 1/(8-3t) ?

The integral from 1 to 2 of 1/(8-3t) is 1/3 (ln(5)-ln(2))

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integral from 1 to 2 of 1/(8-3t)

\int_{1}^{2} \frac{1}{8 - 3 t} d t

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

Decimal Notation

0.30543 \dots

Solution steps

\int_{1}^{2} \frac{1}{8 - 3 t} d t

Apply u-substitution

: \int_{5}^{2} - \frac{1}{3 u} d u

= \int_{5}^{2} - \frac{1}{3 u} d u

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

= - \int_{2}^{5} - \frac{1}{3 u} d u

Take the constant out:

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

= - (- \frac{1}{3} \cdot \int_{2}^{5} \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - (- \frac{1}{3} [ln ∣ u ∣]_{2}^{5})

Simplify

= \frac{1}{3} [ln ∣ u ∣]_{2}^{5}

Compute the boundaries:

ln (5) - ln (2)

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

f (x) = x^{2} - 1, at x = 0

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

\int j

\int (x) sin (x) d x

\frac{d}{d x} (ln (8 - x^{2}))

What is the integral from 1 to 2 of 1/(8-3t) ?
The integral from 1 to 2 of 1/(8-3t) is 1/3 (ln(5)-ln(2))