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

The integral from 1 to 4 of (x^2+4)/x is 15/2+8ln(2)

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integral from 1 to 4 of (x^2+4)/x

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

\frac{15}{2} + 8 ln (2)

Decimal Notation

13.04517 \dots

Solution steps

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

Expand

\frac{x ^{2} + 4}{x} : x + \frac{4}{x}

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

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

\int_{1}^{4} \frac{4}{x} d x = 8 ln (2)

= \frac{15}{2} + 8 ln (2)

\frac{d}{d x} (\frac{x ^{2} + 6}{( 2 x - 7 ) ^{7}})

x \to 0 lim (\frac{200 π}{x})

y = \frac{y + 5 x}{3 y - x}, (1, 2)

\frac{d}{d x} (e^{2 t a n (x)})

1 + t

What is the integral from 1 to 4 of (x^2+4)/x ?
The integral from 1 to 4 of (x^2+4)/x is 15/2+8ln(2)