What is the integral from 1 to 3 of (x^2+4)/x ?

The integral from 1 to 3 of (x^2+4)/x is 4+4ln(3)

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integral from 1 to 3 of (x^2+4)/x

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

4 + 4 ln (3)

Decimal Notation

8.39444 \dots

Solution steps

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

Expand

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

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

\int_{1}^{3} x d x = 4

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

= 4 + 4 ln (3)

\int 20 x^{19} d x

\int_{0}^{1} x a x^{2} + b d x

sin^{2} (θ)

y^{^{'}} = \frac{6 t ^{2}}{y}

\int_{- 1}^{1} x^{2} 5 - 3 x d x

What is the integral from 1 to 3 of (x^2+4)/x ?
The integral from 1 to 3 of (x^2+4)/x is 4+4ln(3)