integral from 0 to x of sqrt(6+7t^2)

解

\int_{0}^{x} 6 + 7 t^{2} d t

\frac{x 7 x ^{2} + 6}{2} + \frac{3 ln \frac{7}{6} x + \frac{7 x ^{2} + 6}{6}}{7}

解答ステップ

\int_{0}^{x} 6 + 7 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{7}{6} x)} \frac{6}{7} 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{6}{7} \cdot \int_{0}^{a r c t a n (\frac{7}{6} x)} sec^{3} (u) d u

積分換算を適用する

= \frac{6}{7} [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{7}{6} x)} + \frac{1}{2} \cdot \int_{0}^{a r c t a n (\frac{7}{6} x)} sec (u) d u

\int_{0}^{a r c t a n (\frac{7}{6} x)} sec (u) d u = ln \frac{7}{6} x + \frac{7 x ^{2} + 6}{6}

= \frac{6}{7} [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{7}{6} x)} + \frac{1}{2} ln \frac{7}{6} x + \frac{7 x ^{2} + 6}{6}

簡素化

\frac{6}{7} [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{7}{6} x)} + \frac{1}{2} ln \frac{7}{6} x + \frac{7 x ^{2} + 6}{6} : \frac{6}{7} [\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (\frac{7}{6} x)} + \frac{1}{2} ln \frac{7}{6} x + \frac{7 x ^{2} + 6}{6}

= \frac{6}{7} [\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (\frac{7}{6} x)} + \frac{1}{2} ln \frac{7}{6} x + \frac{7 x ^{2} + 6}{6}

限度を計算する:

\frac{7}{2 6} x \frac{7 x ^{2} + 6}{6}

= \frac{6}{7} (\frac{7}{2 6} x \frac{7 x ^{2} + 6}{6} + \frac{1}{2} ln \frac{7}{6} x + \frac{7 x ^{2} + 6}{6})

簡素化

= \frac{x 7 x ^{2} + 6}{2} + \frac{3 ln \frac{7}{6} x + \frac{7 x ^{2} + 6}{6}}{7}