What is the integral from 1 to 4 of (x+1/x) ?

The integral from 1 to 4 of (x+1/x) is 15/2+2ln(2)

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integral from 1 to 4 of (x+1/x)

\int_{1}^{4} (x + \frac{1}{x}) d x

\frac{15}{2} + 2 ln (2)

Decimal Notation

8.88629 \dots

Solution steps

\int_{1}^{4} x + \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_{1}^{4} x d x + \int_{1}^{4} \frac{1}{x} d x

\int_{1}^{4} x d x = \frac{15}{2}

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

= \frac{15}{2} + 2 ln (2)

y = x^{2} - \frac{1}{8} cos (x)

(- 1, 4), (11, - 5)

\frac{d}{d x} ((tan (x))^{\frac{7}{x}})

y^{^{'}} - x y = 2 x y

\frac{1}{4} (70 x^{2} - 8 x^{3})

What is the integral from 1 to 4 of (x+1/x) ?
The integral from 1 to 4 of (x+1/x) is 15/2+2ln(2)