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受欢迎的微积分 >

integral of (x+1)sec(x^2+2x)tan(x^2+2x)

解答

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

解答

\frac{1}{2} sec (x^{2} + 2 x) + C

求解步骤

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

使用换元积分法

= \int u sec (u^{2} - 1) tan (u^{2} - 1) d u

使用换元积分法

= \int \frac{sec ( v ) tan ( v )}{2} d v

提出常数:

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

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

使用换元积分法

= \frac{1}{2} \cdot \int 1 d w

常数积分:

\int a d x = a x

= \frac{1}{2} \cdot 1 \cdot w

代回

= \frac{1}{2} \cdot 1 \cdot sec ((x + 1)^{2} - 1)

化简

\frac{1}{2} \cdot 1 \cdot sec ((x + 1)^{2} - 1) : \frac{1}{2} sec (x^{2} + 2 x)

= \frac{1}{2} sec (x^{2} + 2 x)

解答补常数

= \frac{1}{2} sec (x^{2} + 2 x) + C

作图

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