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

The integral from 2 to 4 of (6+u^2)/(u^3) is 9/16+ln(2)

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

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

\frac{9}{16} + ln (2)

Decimal

1.25564 \dots

Solution steps

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

Expand

\frac{6 + u ^{2}}{u ^{3}} : \frac{6}{u ^{3}} + \frac{1}{u}

= \int_{2}^{4} \frac{6}{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_{2}^{4} \frac{6}{u ^{3}} d u + \int_{2}^{4} \frac{1}{u} d u

\int_{2}^{4} \frac{6}{u ^{3}} d u = \frac{9}{16}

\int_{2}^{4} \frac{1}{u} d u = ln (2)

= \frac{9}{16} + ln (2)

x \to 4 lim (2 x + 1)

\frac{d}{d y} (e^{y^{2}})

4 - x^{2}, 0, 2

x \to \frac{π}{2} lim (sin (2 x))

\int \frac{2 x - 5}{x ^{2} + 9} d x

What is the integral from 2 to 4 of (6+u^2)/(u^3) ?
The integral from 2 to 4 of (6+u^2)/(u^3) is 9/16+ln(2)