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受欢迎的微积分 >

integral of 1/(sqrt(16-(2x+2)^2))

解答

\int \frac{1}{16 - ( 2 x + 2 ) ^{2}} d x

解答

\frac{1}{2} arcsin (\frac{1}{2} (x + 1)) + C

求解步骤

\int \frac{1}{16 - ( 2 x + 2 ) ^{2}} d x

16 - (2 x + 2)^{2} = 2 - x^{2} - 2 x + 3

= \int \frac{1}{2 - x ^{2} - 2 x + 3} d x

提出常数:

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

= \frac{1}{2} \cdot \int \frac{1}{- x ^{2} - 2 x + 3} d x

对

- x^{2} - 2 x + 3

配方:

- (x + 1)^{2} + 4

= \frac{1}{2} \cdot \int \frac{1}{- ( x + 1 ) ^{2} + 4} d x

使用换元积分法

= \frac{1}{2} \cdot \int \frac{1}{- u ^{2} + 4} d u

使用三角换元法

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

常数积分:

\int a d x = a x

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

代回

= \frac{1}{2} \cdot 1 \cdot arcsin (\frac{1}{2} (x + 1))

化简

= \frac{1}{2} arcsin (\frac{1}{2} (x + 1))

解答补常数

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

作图

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