What is the integral from 63 to x of 1/(30(35-x)) ?

The integral from 63 to x of 1/(30(35-x)) is -1/30 (ln|35-x|-ln(28))

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integral from 63 to x of 1/(30(35-x))

\int_{63}^{x} \frac{1}{30 ( 35 - x )} d x

- \frac{1}{30} (ln ∣ 35 - x ∣ - ln (28))

Solution steps

\int_{63}^{x} \frac{1}{30 ( 35 - x )} d x

Take the constant out:

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

= \frac{1}{30} \cdot \int_{63}^{x} \frac{1}{35 - x} d x

Apply u-substitution

: \int_{- 28}^{35 - x} - \frac{1}{u} d u

= \frac{1}{30} \cdot \int_{- 28}^{35 - x} - \frac{1}{u} d u

Take the constant out:

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

= \frac{1}{30} (- \int_{- 28}^{35 - x} \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{30} (- [ln ∣ u ∣]_{- 28}^{35 - x})

Simplify

= - \frac{1}{30} [ln ∣ u ∣]_{- 28}^{35 - x}

Compute the boundaries:

ln ∣ 35 - x ∣ - ln (28)

= - \frac{1}{30} (ln ∣ 35 - x ∣ - ln (28))

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What is the integral from 63 to x of 1/(30(35-x)) ?
The integral from 63 to x of 1/(30(35-x)) is -1/30 (ln|35-x|-ln(28))