What is the integral from 1 to 2 of (v^2+4v^6)/(v^3) ?

The integral from 1 to 2 of (v^2+4v^6)/(v^3) is ln(2)+15

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integral from 1 to 2 of (v^2+4v^6)/(v^3)

\int_{1}^{2} \frac{v ^{2} + 4 v ^{6}}{v ^{3}} d v

ln (2) + 15

Decimal Notation

15.69314 \dots

Solution steps

\int_{1}^{2} \frac{v ^{2} + 4 v ^{6}}{v ^{3}} d v

Simplify

\frac{v ^{2} + 4 v ^{6}}{v ^{3}} : \frac{1 + 4 v ^{4}}{v}

= \int_{1}^{2} \frac{1 + 4 v ^{4}}{v} d v

Expand

\frac{1 + 4 v ^{4}}{v} : \frac{1}{v} + 4 v^{3}

= \int_{1}^{2} \frac{1}{v} + 4 v^{3} d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{1}^{2} \frac{1}{v} d v + \int_{1}^{2} 4 v^{3} d v

\int_{1}^{2} \frac{1}{v} d v = ln (2)

\int_{1}^{2} 4 v^{3} d v = 15

= ln (2) + 15

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What is the integral from 1 to 2 of (v^2+4v^6)/(v^3) ?
The integral from 1 to 2 of (v^2+4v^6)/(v^3) is ln(2)+15