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

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

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

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

\frac{3}{2} + ln (2)

Decimal

2.19314 \dots

Solution steps

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

Expand

\frac{4 + u ^{2}}{u ^{3}} : \frac{4}{u ^{3}} + \frac{1}{u}

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

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

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

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

= \frac{3}{2} + ln (2)

\frac{d y}{d x} = \frac{x ^{2}}{y}

\frac{d x}{d t} = 2 (x^{2} + 1), x (\frac{π}{4}) = 1

\int \frac{x ^{2} - 2 x - 1}{x} d x

e^{1 - t}

x y^{^{'}} sin^{2} (\frac{y}{x}) = x + y sin^{2} (\frac{y}{x})

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