What is the integral from b to 1 of /r ?

The integral from b to 1 of /r is ln|d|-ln|b|

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integral from b to 1 of /r

\int_{b}^{d} \frac{1}{r} d r

ln ∣ d ∣ - ln ∣ b ∣

Solution steps

\int_{b}^{d} \frac{1}{r} d r

Use the common integral:

\int \frac{1}{r} d r = ln (∣ r ∣)

= [ln ∣ r ∣]_{b}^{d}

Compute the boundaries:

ln ∣ d ∣ - ln ∣ b ∣

= ln ∣ d ∣ - ln ∣ b ∣

f (x) = 7 x^{3} + 5 x, at x = 7

\frac{d}{d x} (3 e^{x} (5 x^{5} - 2))

y^{'} = sin^{2} (x - y)

\frac{d}{d x} ((x^{2} + 3)^{4})

\frac{d}{d x} (5 e^{x})

What is the integral from b to 1 of /r ?
The integral from b to 1 of /r is ln|d|-ln|b|