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

The integral from 4 to 7 of 1/(x-3) is 2ln(2)

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

\int_{4}^{7} \frac{1}{x - 3} d x

2 ln (2)

Decimal Notation

1.38629 \dots

Solution steps

\int_{4}^{7} \frac{1}{x - 3} d x

Apply u-substitution

: \int_{1}^{4} \frac{1}{u} d u

= \int_{1}^{4} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{1}^{4}

Compute the boundaries:

2 ln (2)

= 2 ln (2)

\int - \frac{4}{3} x^{3} 2 x - 1 d x

ln (x^{2} + 1)

\int 2 x sin (x) e^{x} d x

\frac{e ^{- 4 s}}{2 s - 1}

What is the integral from 4 to 7 of 1/(x-3) ?
The integral from 4 to 7 of 1/(x-3) is 2ln(2)