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

解

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

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

十進法表記

2.36150 \dots

解答ステップ

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

拡張

(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

総和規則を適用する:

\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

分数を組み合わせる

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

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