What is the integral from-1 to 1 of (12-4x)^{-1} ?

The integral from-1 to 1 of (12-4x)^{-1} is 1/4 ln(2)

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integral from-1 to 1 of (12-4x)^{-1}

\int_{- 1}^{1} (12 - 4 x)^{- 1} d x

\frac{1}{4} ln (2)

Decimal Notation

0.17328 \dots

Solution steps

\int_{- 1}^{1} (12 - 4 x)^{- 1} d x

Apply exponent rule:

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

= \int_{- 1}^{1} \frac{1}{12 - 4 x} d x

Apply u-substitution

: \int_{16}^{8} - \frac{1}{4 u} d u

= \int_{16}^{8} - \frac{1}{4 u} d u

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

= - \int_{8}^{16} - \frac{1}{4 u} d u

Take the constant out:

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

= - (- \frac{1}{4} \cdot \int_{8}^{16} \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - (- \frac{1}{4} [ln ∣ u ∣]_{8}^{16})

Simplify

= \frac{1}{4} [ln ∣ u ∣]_{8}^{16}

Compute the boundaries:

ln (2)

= \frac{1}{4} ln (2)

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e^{- x^{3}}

What is the integral from-1 to 1 of (12-4x)^{-1} ?
The integral from-1 to 1 of (12-4x)^{-1} is 1/4 ln(2)