What is the integral from v^1 to v^2 of 1/v ?

The integral from v^1 to v^2 of 1/v is 2ln(v)-ln|v|

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integral from v^1 to v^2 of 1/v

\int_{v^{1}}^{v^{2}} \frac{1}{v} d v

2 ln (v) - ln ∣ v ∣

Solution steps

\int_{v^{1}}^{v^{2}} \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= [ln ∣ v ∣]_{v^{1}}^{v^{2}}

Simplify

= [ln ∣ v ∣]_{v}^{v^{2}}

Compute the boundaries:

2 ln (v) - ln ∣ v ∣

= 2 ln (v) - ln ∣ v ∣

x \to - 1 lim (\frac{1}{x ^{2} - 1})

\frac{d ^{2}}{d x ^{2}} (9 x sin (x^{2}))

\frac{\partial}{\partial x} (2 x + 1)

\frac{\partial}{\partial z} (z cos (x z))

\int_{1}^{2} 2 r^{2} ln (r) d r

What is the integral from v^1 to v^2 of 1/v ?
The integral from v^1 to v^2 of 1/v is 2ln(v)-ln|v|