What is the integral from-1 to 1 of 1/(3x-4) ?

The integral from-1 to 1 of 1/(3x-4) is -1/3 ln(7)

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integral from-1 to 1 of 1/(3x-4)

\int_{- 1}^{1} \frac{1}{3 x - 4} d x

- \frac{1}{3} ln (7)

Decimal

- 0.64863 \dots

Solution steps

\int_{- 1}^{1} \frac{1}{3 x - 4} d x

Apply u-substitution

: \int_{- 7}^{- 1} \frac{1}{3 u} d u

= \int_{- 7}^{- 1} \frac{1}{3 u} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int_{- 7}^{- 1} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} [ln ∣ u ∣]_{- 7}^{- 1}

Compute the boundaries:

- ln (7)

= \frac{1}{3} (- ln (7))

Simplify

= - \frac{1}{3} ln (7)

\int_{e}^{x} sin (t) d t

\int_{0}^{1.2} 3 3 x + 4 d x

\int_{0}^{\frac{1}{2}} x (1 - 4 x^{2})^{10} d x

\int_{2}^{\infty} t e^{- s t} d t

\int_{- 1}^{0} 3 x + 4 d x

What is the integral from-1 to 1 of 1/(3x-4) ?
The integral from-1 to 1 of 1/(3x-4) is -1/3 ln(7)