What is the integral from-1 to 1 of (13-12u)^{-1} ?

The integral from-1 to 1 of (13-12u)^{-1} is 1/6 ln(5)

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integral from-1 to 1 of (13-12u)^{-1}

\int_{- 1}^{1} (13 - 12 u)^{- 1} d u

\frac{1}{6} ln (5)

Solution steps

\int_{- 1}^{1} (13 - 12 u)^{- 1} d u

Apply exponent rule:

a^{- 1} = \frac{1}{a}

= \int_{- 1}^{1} \frac{1}{13 - 12 u} d u

Apply u-substitution

: \int_{25}^{1} - \frac{1}{12 v} d v

= \int_{25}^{1} - \frac{1}{12 v} d v

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

= - \int_{1}^{25} - \frac{1}{12 v} d v

Take the constant out:

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

= - (- \frac{1}{12} \cdot \int_{1}^{25} \frac{1}{v} d v)

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= - (- \frac{1}{12} [ln ∣ v ∣]_{1}^{25})

Simplify

= \frac{1}{12} [ln ∣ v ∣]_{1}^{25}

Compute the boundaries:

2 ln (5)

= \frac{1}{12} \cdot 2 ln (5)

Simplify

= \frac{1}{6} ln (5)

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What is the integral from-1 to 1 of (13-12u)^{-1} ?
The integral from-1 to 1 of (13-12u)^{-1} is 1/6 ln(5)