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

The integral from-4 to-3 of 1/(x-3) is ln(6)-ln(7)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

ln (6) - ln (7)

Decimal

- 0.15415 \dots

Solution steps

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

Apply u-substitution

: \int_{- 7}^{- 6} \frac{1}{u} d u

= \int_{- 7}^{- 6} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{- 7}^{- 6}

Compute the boundaries:

ln (6) - ln (7)

= ln (6) - ln (7)

\int_{- 1}^{216} x^{- \frac{1}{3}} d x

\int_{1}^{- 1} (6 x + 5)^{2} d x

\int_{0}^{0.5} ln (x + 1) d x

\int_{0}^{2} (3 t - 1)^{68} d t

\int_{0}^{1} x e^{- 8 x} d x

What is the integral from-4 to-3 of 1/(x-3) ?
The integral from-4 to-3 of 1/(x-3) is ln(6)-ln(7)