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

The integral from 1 to 4 of 1/(4-x) is diverges

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

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

diverges

Solution steps

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

Apply u-substitution

: \int_{3}^{0} - \frac{1}{u} d u

= \int_{3}^{0} - \frac{1}{u} d u

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

= - \int_{0}^{3} - \frac{1}{u} d u

Take the constant out:

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

= - (- \int_{0}^{3} \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - (- [ln ∣ u ∣]_{0}^{3})

Simplify

= [ln ∣ u ∣]_{0}^{3}

Compute the boundaries:

diverges

= diverges

\int_{- 5}^{5} 3 x^{2} - 6 x - 9 d x

\int_{0}^{1} \frac{13}{4} + \frac{9}{4} x d x

\int_{0}^{1} 1 - r^{2} r d r

\int_{1}^{2} 4 x^{3} - 3 x^{2} + 4 x + 2 d x

\int_{3}^{x} (t^{2} + 1) d t

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