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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 4 to 6 of 1/(x-3)

\int_{4}^{6} \frac{1}{x - 3} d x

ln (3)

Decimal

1.09861 \dots

Solution steps

\int_{4}^{6} \frac{1}{x - 3} d x

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{1}^{3}

Compute the boundaries:

ln (3)

= ln (3)

\int_{1}^{5} \frac{1}{1 + x ^{3}} d x

\int_{- 3}^{3} (5 x^{3} + 2 x^{2} + 6 x - 1) d x

\int_{0}^{\frac{π}{2}} cos (x) cos (n x) d x

\int_{1}^{x} 2 - t d t

\int_{- π}^{\frac{π}{2}} cos (n x) d x

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