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

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

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

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

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

Decimal

1.25564 \dots

Solution steps

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

Expand

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

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

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

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

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

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

y = ln (9 x^{2} + 5 y^{2})

x \to \infty lim (ln (7 x + 1) - ln (4 x + 1))

\int_{0}^{5} 25 - x^{2} d x

\int 5 e^{x} - \frac{2}{x} d x

\frac{d}{d x} (3 x + 2 e^{x})

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