What is the integral from-1 to 1 of (14-12x)^{-1} ?

The integral from-1 to 1 of (14-12x)^{-1} is 1/12 ln(13)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from-1 to 1 of (14-12x)^{-1}

\int_{- 1}^{1} (14 - 12 x)^{- 1} d x

\frac{1}{12} ln (13)

Decimal Notation

0.21374 \dots

Solution steps

\int_{- 1}^{1} (14 - 12 x)^{- 1} d x

Apply exponent rule:

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

= \int_{- 1}^{1} \frac{1}{14 - 12 x} d x

Apply u-substitution

: \int_{26}^{2} - \frac{1}{12 u} d u

= \int_{26}^{2} - \frac{1}{12 u} d u

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

= - \int_{2}^{26} - \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_{2}^{26} \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - (- \frac{1}{12} [ln ∣ u ∣]_{2}^{26})

Simplify

= \frac{1}{12} [ln ∣ u ∣]_{2}^{26}

Compute the boundaries:

ln (26) - ln (2)

= \frac{1}{12} (ln (26) - ln (2))

Simplify

= \frac{1}{12} ln (13)

\frac{\partial}{\partial z} (y arctan (\frac{x}{z}))

\int (7 + 2 x) e^{x} d x

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

\int_{0}^{4} 1 + \frac{9 x}{4} d x

\int_{0}^{π} sin (4 x) d x

What is the integral from-1 to 1 of (14-12x)^{-1} ?
The integral from-1 to 1 of (14-12x)^{-1} is 1/12 ln(13)