What is the integral from 1 to 4 of (x/8-3/x) ?

The integral from 1 to 4 of (x/8-3/x) is 15/16-6ln(2)

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integral from 1 to 4 of (x/8-3/x)

\int_{1}^{4} (\frac{x}{8} - \frac{3}{x}) d x

\frac{15}{16} - 6 ln (2)

Decimal

- 3.22138 \dots

Solution steps

\int_{1}^{4} \frac{x}{8} - \frac{3}{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_{1}^{4} \frac{x}{8} d x - \int_{1}^{4} \frac{3}{x} d x

\int_{1}^{4} \frac{x}{8} d x = \frac{15}{16}

\int_{1}^{4} \frac{3}{x} d x = 6 ln (2)

= \frac{15}{16} - 6 ln (2)

\int_{0}^{1} \frac{( - x + 1 ) ^{3}}{6} d x

\int_{1}^{3} h (x) d x

\int_{1}^{3} x e^{2 x} d x

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

\int_{0}^{1} e^{6 y} y d y

What is the integral from 1 to 4 of (x/8-3/x) ?
The integral from 1 to 4 of (x/8-3/x) is 15/16-6ln(2)