What is the integral from 1 to 2 of (v^3+5v^8)/(v^4) ?

The integral from 1 to 2 of (v^3+5v^8)/(v^4) is ln(2)+31

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integral from 1 to 2 of (v^3+5v^8)/(v^4)

\int_{1}^{2} \frac{v ^{3} + 5 v ^{8}}{v ^{4}} d v

ln (2) + 31

Decimal

31.69314 \dots

Solution steps

\int_{1}^{2} \frac{v ^{3} + 5 v ^{8}}{v ^{4}} d v

Simplify

\frac{v ^{3} + 5 v ^{8}}{v ^{4}} : \frac{1 + 5 v ^{5}}{v}

= \int_{1}^{2} \frac{1 + 5 v ^{5}}{v} d v

Expand

\frac{1 + 5 v ^{5}}{v} : \frac{1}{v} + 5 v^{4}

= \int_{1}^{2} \frac{1}{v} + 5 v^{4} 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} 5 v^{4} d v

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

\int_{1}^{2} 5 v^{4} d v = 31

= ln (2) + 31

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What is the integral from 1 to 2 of (v^3+5v^8)/(v^4) ?
The integral from 1 to 2 of (v^3+5v^8)/(v^4) is ln(2)+31