integral from 0 to u of sqrt(11+12t^2)

解答

\int_{0}^{u} 11 + 12 t^{2} d t

\frac{u 12 u ^{2} + 11}{2} + \frac{11}{4 3} ln \frac{2 3 u}{11} + \frac{12 u ^{2} + 11}{11}

求解步骤

\int_{0}^{u} 11 + 12 t^{2} d t

使用三角换元法

= \int_{0}^{a r c t a n (\frac{2 3 u}{11})} \frac{11}{2 3} sec^{3} (v) d v

提出常数:

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

= \frac{11}{2 3} \cdot \int_{0}^{a r c t a n (\frac{2 3 u}{11})} sec^{3} (v) d v

使用降次积分法

= \frac{11}{2 3} [\frac{sec ^{2} ( v ) sin ( v )}{2}]_{0}^{a r c t a n (\frac{2 3 u}{11})} + \frac{1}{2} \cdot \int_{0}^{a r c t a n (\frac{2 3 u}{11})} sec (v) d v

\int_{0}^{a r c t a n (\frac{2 3 u}{11})} sec (v) d v = ln \frac{2 3 u}{11} + \frac{12 u ^{2} + 11}{11}

= \frac{11}{2 3} [\frac{sec ^{2} ( v ) sin ( v )}{2}]_{0}^{a r c t a n (\frac{2 3 u}{11})} + \frac{1}{2} ln \frac{2 3 u}{11} + \frac{12 u ^{2} + 11}{11}

化简

\frac{11}{2 3} [\frac{sec ^{2} ( v ) sin ( v )}{2}]_{0}^{a r c t a n (\frac{2 3 u}{11})} + \frac{1}{2} ln \frac{2 3 u}{11} + \frac{12 u ^{2} + 11}{11} : \frac{11}{2 3} [\frac{1}{2} sec (v) tan (v)]_{0}^{a r c t a n (\frac{2 3 u}{11})} + \frac{1}{2} ln \frac{2 3 u}{11} + \frac{12 u ^{2} + 11}{11}

= \frac{11}{2 3} [\frac{1}{2} sec (v) tan (v)]_{0}^{a r c t a n (\frac{2 3 u}{11})} + \frac{1}{2} ln \frac{2 3 u}{11} + \frac{12 u ^{2} + 11}{11}

带入上下限计算后得:

\frac{3 u 12 u ^{2} + 11}{11}

= \frac{11}{2 3} (\frac{3 u 12 u ^{2} + 11}{11} + \frac{1}{2} ln \frac{2 3 u}{11} + \frac{12 u ^{2} + 11}{11})

化简

= \frac{u 12 u ^{2} + 11}{2} + \frac{11}{4 3} ln \frac{2 3 u}{11} + \frac{12 u ^{2} + 11}{11}

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