integral from 0 to 1 of (1+x)/(1+x^2)

解

\int_{0}^{1} \frac{1 + x}{1 + x ^{2}} d x

\frac{π}{4} + \frac{1}{2} ln (2)

十進法表記

1.13197 \dots

解答ステップ

\int_{0}^{1} \frac{1 + x}{1 + x ^{2}} d x

拡張

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

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_{0}^{1} \frac{1}{1 + x ^{2}} d x + \int_{0}^{1} \frac{x}{1 + x ^{2}} d x

\int_{0}^{1} \frac{1}{1 + x ^{2}} d x = \frac{π}{4}

\int_{0}^{1} \frac{x}{1 + x ^{2}} d x = \frac{1}{2} ln (2)

= \frac{π}{4} + \frac{1}{2} ln (2)