integral of sqrt(2+4t^2)

解答

\int 2 + 4 t^{2} d t

\frac{1}{2} t 1 + 2 t^{2} + \frac{1}{2} ln 2 t + 1 + 2 t^{2} + C

求解步骤

\int 2 + 4 t^{2} d t

使用三角换元法

= \int sec^{3} (u) d u

使用降次积分法

= \frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} \cdot \int sec (u) d u

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= \frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣

u = arctan (\frac{2}{2} t)

代回

= \frac{sec ^{2} ( arctan ( \frac{2}{2} t ) ) sin ( arctan ( \frac{2}{2} t ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{2}{2} t)) + sec (arctan (\frac{2}{2} t))

化简

\frac{sec ^{2} ( arctan ( \frac{2}{2} t ) ) sin ( arctan ( \frac{2}{2} t ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{2}{2} t)) + sec (arctan (\frac{2}{2} t)) : \frac{1}{2} t 1 + 2 t^{2} + \frac{1}{2} ln 2 t + 1 + 2 t^{2}

= \frac{1}{2} t 1 + 2 t^{2} + \frac{1}{2} ln 2 t + 1 + 2 t^{2}

解答补常数

= \frac{1}{2} t 1 + 2 t^{2} + \frac{1}{2} ln 2 t + 1 + 2 t^{2} + C

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