integral from 1 to 2 of ((1+t^2)/(t^4))

解

\int_{1}^{2} (\frac{1 + t ^{2}}{t ^{4}}) d t

\frac{19}{24}

十進法表記

0.79166 \dots

解答ステップ

\int_{1}^{2} \frac{1 + t ^{2}}{t ^{4}} d t

拡張

\frac{1 + t ^{2}}{t ^{4}} : \frac{1}{t ^{4}} + \frac{1}{t ^{2}}

= \int_{1}^{2} \frac{1}{t ^{4}} + \frac{1}{t ^{2}} d t

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}{t ^{4}} d t + \int_{1}^{2} \frac{1}{t ^{2}} d t

\int_{1}^{2} \frac{1}{t ^{4}} d t = \frac{7}{24}

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

= \frac{7}{24} + \frac{1}{2}

簡素化

\frac{7}{24} + \frac{1}{2} : \frac{19}{24}

= \frac{19}{24}