integral from 1 to 2 of (m^2+1)/(m^3)

解

\int_{1}^{2} \frac{m ^{2} + 1}{m ^{3}} d m

ln (2) + \frac{3}{8}

十進法表記

1.06814 \dots

解答ステップ

\int_{1}^{2} \frac{m ^{2} + 1}{m ^{3}} d m

拡張

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

= \int_{1}^{2} \frac{1}{m} + \frac{1}{m ^{3}} d m

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} \frac{1}{m} d m + \int_{1}^{2} \frac{1}{m ^{3}} d m

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

\int_{1}^{2} \frac{1}{m ^{3}} d m = \frac{3}{8}

= ln (2) + \frac{3}{8}