What is the integral from 0 to 1 of 3/(2t-5) ?

The integral from 0 to 1 of 3/(2t-5) is 3/2 (ln(3)-ln(5))

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integral from 0 to 1 of 3/(2t-5)

\int_{0}^{1} \frac{3}{2 t - 5} d t

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

Decimal

- 0.76623 \dots

Solution steps

\int_{0}^{1} \frac{3}{2 t - 5} d t

Take the constant out:

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

= 3 \cdot \int_{0}^{1} \frac{1}{2 t - 5} d t

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

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

Simplify

3 \cdot \frac{1}{2} [ln ∣ u ∣]_{- 5}^{- 3} : \frac{3}{2} [ln ∣ u ∣]_{- 5}^{- 3}

= \frac{3}{2} [ln ∣ u ∣]_{- 5}^{- 3}

Compute the boundaries:

ln (3) - ln (5)

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

\int_{1}^{4} x^{2} - 4 x - 3 d x

\int_{- \infty}^{4} x e^{x} d x

\frac{3}{s ^{2} + 3 s + 6}

\frac{2 x ^{4}}{b ^{2} - x ^{2}}

\int \frac{( x ^{2} - x + 24 )}{x ^{3} + 6 x} d x

What is the integral from 0 to 1 of 3/(2t-5) ?
The integral from 0 to 1 of 3/(2t-5) is 3/2 (ln(3)-ln(5))