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受欢迎的微积分 >

integral of 1/(sqrt(2x^2+4x+7))

解答

\int \frac{1}{2 x ^{2} + 4 x + 7} d x

解答

\frac{1}{2} ln \frac{2}{5} (x + 1) + \frac{2 x ^{2} + 4 x + 7}{5} + C

求解步骤

\int \frac{1}{2 x ^{2} + 4 x + 7} d x

对

2 x^{2} + 4 x + 7

配方:

2 (x + 1)^{2} + 5

= \int \frac{1}{2 ( x + 1 ) ^{2} + 5} d x

使用换元积分法

= \int \frac{1}{2 u ^{2} + 5} d u

使用三角换元法

= \int \frac{sec ( 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) d v

使用积分常见定则:

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

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

代回

= \frac{1}{2} ln tan (arctan (\frac{2}{5} (x + 1))) + sec (arctan (\frac{2}{5} (x + 1)))

化简

\frac{1}{2} ln tan (arctan (\frac{2}{5} (x + 1))) + sec (arctan (\frac{2}{5} (x + 1))) : \frac{1}{2} ln \frac{2}{5} (x + 1) + \frac{2 x ^{2} + 4 x + 7}{5}

= \frac{1}{2} ln \frac{2}{5} (x + 1) + \frac{2 x ^{2} + 4 x + 7}{5}

解答补常数

= \frac{1}{2} ln \frac{2}{5} (x + 1) + \frac{2 x ^{2} + 4 x + 7}{5} + C

作图

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