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

解

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

2 ln (2) + \frac{3}{2}

十進法表記

2.88629 \dots

解答ステップ

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

拡張

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

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

\int_{1}^{2} 1 d x = 1

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

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

= 1 + 2 ln (2) + \frac{1}{2}

分数を組み合わせる

1 + \frac{1}{2} : \frac{3}{2}

= 2 ln (2) + \frac{3}{2}