What is the integral from-1 to 1 of (20-12x)^{-1} ?

The integral from-1 to 1 of (20-12x)^{-1} is 1/6 ln(2)

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integral from-1 to 1 of (20-12x)^{-1}

\int_{- 1}^{1} (20 - 12 x)^{- 1} d x

\frac{1}{6} ln (2)

Decimal Notation

0.11552 \dots

Solution steps

\int_{- 1}^{1} (20 - 12 x)^{- 1} d x

Apply exponent rule:

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

= \int_{- 1}^{1} \frac{1}{20 - 12 x} d x

Apply u-substitution

: \int_{32}^{8} - \frac{1}{12 u} d u

= \int_{32}^{8} - \frac{1}{12 u} d u

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

= - \int_{8}^{32} - \frac{1}{12 u} d u

Take the constant out:

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

= - (- \frac{1}{12} \cdot \int_{8}^{32} \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - (- \frac{1}{12} [ln ∣ u ∣]_{8}^{32})

Simplify

= \frac{1}{12} [ln ∣ u ∣]_{8}^{32}

Compute the boundaries:

2 ln (2)

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

Simplify

= \frac{1}{6} ln (2)

\int \frac{( sin ( 2 x ) )}{e ^{x}} d x

\int \frac{( x ^{2} + 5 x + 1 )}{( x ^{2} + 1 ) ^{2}} d x

x (4 - x)^{3}

\frac{d}{d t} (e^{t^{3}})

y = \frac{4}{x} + x + 0.5 - 5 x

What is the integral from-1 to 1 of (20-12x)^{-1} ?
The integral from-1 to 1 of (20-12x)^{-1} is 1/6 ln(2)