integral from 0 to x of sqrt(13+14t^2)

解

\int_{0}^{x} 13 + 14 t^{2} d t

\frac{13}{14} (\frac{7}{26} x \frac{14 x ^{2} + 13}{13} + \frac{1}{2} ln \frac{14}{13} x + \frac{14 x ^{2} + 13}{13})

解答ステップ

\int_{0}^{x} 13 + 14 t^{2} d t

三角関数による置換を適用する

= \int_{0}^{a r c t a n (\frac{14}{13} x)} \frac{13}{14} sec^{3} (u) d u

定数を除く:

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

= \frac{13}{14} \cdot \int_{0}^{a r c t a n (\frac{14}{13} x)} sec^{3} (u) d u

積分換算を適用する

= \frac{13}{14} [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{14}{13} x)} + \frac{1}{2} \cdot \int_{0}^{a r c t a n (\frac{14}{13} x)} sec (u) d u

\int_{0}^{a r c t a n (\frac{14}{13} x)} sec (u) d u = ln \frac{14}{13} x + \frac{14 x ^{2} + 13}{13}

= \frac{13}{14} [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{14}{13} x)} + \frac{1}{2} ln \frac{14}{13} x + \frac{14 x ^{2} + 13}{13}

簡素化

\frac{13}{14} [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{14}{13} x)} + \frac{1}{2} ln \frac{14}{13} x + \frac{14 x ^{2} + 13}{13} : \frac{13}{14} [\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (\frac{14}{13} x)} + \frac{1}{2} ln \frac{14}{13} x + \frac{14 x ^{2} + 13}{13}

= \frac{13}{14} [\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (\frac{14}{13} x)} + \frac{1}{2} ln \frac{14}{13} x + \frac{14 x ^{2} + 13}{13}

限度を計算する:

\frac{7}{26} x \frac{14 x ^{2} + 13}{13}

= \frac{13}{14} (\frac{7}{26} x \frac{14 x ^{2} + 13}{13} + \frac{1}{2} ln \frac{14}{13} x + \frac{14 x ^{2} + 13}{13})