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

The integral from 1 to 4 of 4/(1-3x) is 4/3 (ln(2)-ln(11))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{4}{3} (ln (2) - ln (11))

Decimal Notation

- 2.27299 \dots

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int_{- 2}^{- 11} - \frac{1}{3 u} d u

= 4 \cdot \int_{- 2}^{- 11} - \frac{1}{3 u} d u

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

= 4 (- \int_{- 11}^{- 2} - \frac{1}{3 u} d u)

Take the constant out:

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

= 4 (- (- \frac{1}{3} \cdot \int_{- 11}^{- 2} \frac{1}{u} d u))

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 4 (- (- \frac{1}{3} [ln ∣ u ∣]_{- 11}^{- 2}))

Simplify

4 (- (- \frac{1}{3} [ln ∣ u ∣]_{- 11}^{- 2})) : \frac{4}{3} [ln ∣ u ∣]_{- 11}^{- 2}

= \frac{4}{3} [ln ∣ u ∣]_{- 11}^{- 2}

Compute the boundaries:

ln (2) - ln (11)

= \frac{4}{3} (ln (2) - ln (11))

\int_{0}^{1} 2 t e^{t^{2} - 1} + 3 t^{7} d t

\int_{- 1}^{3} x^{2} + 2 x + 1 d x

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

\int_{2 x}^{x^{2}} t d t

\int_{0}^{\frac{1}{5}} 4 x d x

What is the integral from 1 to 4 of 4/(1-3x) ?
The integral from 1 to 4 of 4/(1-3x) is 4/3 (ln(2)-ln(11))