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

The integral from 1 to e of ((x^2+1))/x is (e^2-1)/2+1

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

\int_{1}^{e} \frac{( x ^{2} + 1 )}{x} d x

\frac{e ^{2} - 1}{2} + 1

Decimal Notation

4.19452 \dots

Solution steps

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

Expand

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

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

\int_{1}^{e} x d x = \frac{e ^{2} - 1}{2}

\int_{1}^{e} \frac{1}{x} d x = 1

= \frac{e ^{2} - 1}{2} + 1

x \to 4 + lim (ln (x^{2} - 16))

\frac{1}{2} y^{^{'}} + y = 0, y (1) = 1

f (x) = 1 - x^{2}, (1, 0)

\frac{d}{d x} (\frac{( 6 x ^{2} + 6 x + 8 )}{x})

\frac{\partial}{\partial y} (x 1 + y^{3})

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