What is the integral from 3 to 5 of (x^2+1)/x ?

The integral from 3 to 5 of (x^2+1)/x is 8+ln(5)-ln(3)

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integral from 3 to 5 of (x^2+1)/x

\int_{3}^{5} \frac{x ^{2} + 1}{x} d x

8 + ln (5) - ln (3)

Decimal Notation

8.51082 \dots

Solution steps

\int_{3}^{5} \frac{x ^{2} + 1}{x} d x

Expand

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

= \int_{3}^{5} 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_{3}^{5} x d x + \int_{3}^{5} \frac{1}{x} d x

\int_{3}^{5} x d x = 8

\int_{3}^{5} \frac{1}{x} d x = ln (5) - ln (3)

= 8 + ln (5) - ln (3)

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\frac{\partial}{\partial y} ((y e^{x})^{2})

f (x) = (x^{- 2} + x)^{3} (4 x)^{- 2 x}

What is the integral from 3 to 5 of (x^2+1)/x ?
The integral from 3 to 5 of (x^2+1)/x is 8+ln(5)-ln(3)