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

解

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

ln (3) - ln (2) - \frac{1}{6}

十進法表記

0.23879 \dots

解答ステップ

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

u置換積分法を適用する

= \int_{2}^{3} \frac{u - 1}{u ^{2}} d u

拡張

\frac{u - 1}{u ^{2}} : \frac{1}{u} - \frac{1}{u ^{2}}

= \int_{2}^{3} \frac{1}{u} - \frac{1}{u ^{2}} d u

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_{2}^{3} \frac{1}{u} d u - \int_{2}^{3} \frac{1}{u ^{2}} d u

\int_{2}^{3} \frac{1}{u} d u = ln (3) - ln (2)

\int_{2}^{3} \frac{1}{u ^{2}} d u = \frac{1}{6}

= ln (3) - ln (2) - \frac{1}{6}