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

The integral from 1 to e of (x+1/(x^2))ln(x) is (e^3+e-8)/(4e)+1

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

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

\frac{e ^{3} + e - 8}{4 e} + 1

Decimal Notation

2.36150 \dots

Solution steps

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

Expand

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

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

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

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

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

Combine the fractions

\frac{e ^{2} + 1}{4} - \frac{2}{e} : \frac{e ^{3} + e - 8}{4 e}

= \frac{e ^{3} + e - 8}{4 e} + 1

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What is the integral from 1 to e of (x+1/(x^2))ln(x) ?
The integral from 1 to e of (x+1/(x^2))ln(x) is (e^3+e-8)/(4e)+1