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

The integral from 1 to 2 of (v^3+3v^6)/(v^4) is ln(2)+7

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

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

ln (2) + 7

Decimal

7.69314 \dots

Solution steps

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

Expand

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

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

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

\int_{1}^{2} 3 v^{2} d v = 7

= ln (2) + 7

y = x^{2} - 2 x, y = 4 x + 7

f (x) = \frac{x ^{2}}{x - 4}

lo g_{3} (x) - 2

\frac{\partial}{\partial y} (e^{x} x y + e^{x} y^{2} + e^{x} y)

\int_{0}^{1} \frac{2 x}{( 2 - x ) ^{2}} d x

What is the integral from 1 to 2 of (v^3+3v^6)/(v^4) ?
The integral from 1 to 2 of (v^3+3v^6)/(v^4) is ln(2)+7