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

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

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

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

\frac{e ^{2} + 1}{4} + \frac{1}{2}

Decimal Notation

2.59726 \dots

Solution steps

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

Expand

\frac{( x ^{2} + 1 ) ln ( x )}{x} : x ln (x) + \frac{ln ( x )}{x}

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

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

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

= \frac{e ^{2} + 1}{4} + \frac{1}{2}

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