integral from 0 to e^w of sqrt(1+e^w)

解

\int_{0}^{e^{w}} 1 + e^{w} d w

- ln 1 + e^{e^{w}} + 1 + ln 1 + e^{e^{w}} - 1 + ln (2 + 1) - ln (2 - 1) + 2 1 + e^{e^{w}} - 22

解答ステップ

\int_{0}^{e^{w}} 1 + e^{w} d w

u置換積分法を適用する

= \int_{1}^{e^{e^{w}}} \frac{1 + u}{u} d u

u置換積分法を適用する

= \int_{2}^{1 + e^{e^{w}}} \frac{2 v ^{2}}{v ^{2} - 1} d v

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int_{2}^{1 + e^{e^{w}}} \frac{v ^{2}}{v ^{2} - 1} d v

因数

v^{2} - 1 : - (- v^{2} + 1)

= 2 \cdot \int_{2}^{1 + e^{e^{w}}} \frac{v ^{2}}{- ( - v ^{2} + 1 )} d v

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 (- \int_{2}^{1 + e^{e^{w}}} \frac{v ^{2}}{- v ^{2} + 1} d v)

\frac{v ^{2}}{- v ^{2} + 1} = \frac{1}{- v ^{2} + 1} - 1

= 2 (- \int_{2}^{1 + e^{e^{w}}} \frac{1}{- v ^{2} + 1} - 1 d v)

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

= 2 (- (\int_{2}^{1 + e^{e^{w}}} \frac{1}{- v ^{2} + 1} d v - \int_{2}^{1 + e^{e^{w}}} 1 d v))

\int_{2}^{1 + e^{e^{w}}} \frac{1}{- v ^{2} + 1} d v = \frac{1}{2} (ln 1 + e^{e^{w}} + 1 - ln 1 + e^{e^{w}} - 1) - \frac{1}{2} (ln (2 + 1) - ln (2 - 1))

\int_{2}^{1 + e^{e^{w}}} 1 d v = 1 + e^{e^{w}} - 2

= 2 (- (\frac{1}{2} (ln 1 + e^{e^{w}} + 1 - ln 1 + e^{e^{w}} - 1) - \frac{1}{2} (ln (2 + 1) - ln (2 - 1)) - (1 + e^{e^{w}} - 2)))

簡素化

2 (- (\frac{1}{2} (ln 1 + e^{e^{w}} + 1 - ln 1 + e^{e^{w}} - 1) - \frac{1}{2} (ln (2 + 1) - ln (2 - 1)) - (1 + e^{e^{w}} - 2))) : - ln 1 + e^{e^{w}} + 1 + ln 1 + e^{e^{w}} - 1 + ln (2 + 1) - ln (2 - 1) + 2 1 + e^{e^{w}} - 22

= - ln 1 + e^{e^{w}} + 1 + ln 1 + e^{e^{w}} - 1 + ln (2 + 1) - ln (2 - 1) + 2 1 + e^{e^{w}} - 22