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受欢迎的微积分 >

integral of xsec(3x^2+1)

解答

\int x sec (3 x^{2} + 1) d x

解答

\frac{1}{6} ln tan (3 x^{2} + 1) + sec (3 x^{2} + 1) + C

求解步骤

\int x sec (3 x^{2} + 1) d x

使用换元积分法

= \int \frac{sec ( 3 u + 1 )}{2} d u

提出常数:

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

= \frac{1}{2} \cdot \int sec (3 u + 1) d u

使用换元积分法

= \frac{1}{2} \cdot \int sec (v) \frac{1}{3} d v

提出常数:

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

= \frac{1}{2} \cdot \frac{1}{3} \cdot \int sec (v) d v

使用积分常见定则:

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

= \frac{1}{2} \cdot \frac{1}{3} ln ∣ tan (v) + sec (v) ∣

代回

= \frac{1}{2} \cdot \frac{1}{3} ln tan (3 x^{2} + 1) + sec (3 x^{2} + 1)

化简

\frac{1}{2} \cdot \frac{1}{3} ln tan (3 x^{2} + 1) + sec (3 x^{2} + 1) : \frac{1}{6} ln tan (3 x^{2} + 1) + sec (3 x^{2} + 1)

= \frac{1}{6} ln tan (3 x^{2} + 1) + sec (3 x^{2} + 1)

解答补常数

= \frac{1}{6} ln tan (3 x^{2} + 1) + sec (3 x^{2} + 1) + C

作图

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