integral of (-6)/(sqrt(x^2+16))

解答

\int \frac{- 6}{x ^{2} + 16} d x

- 6 ln (\frac{1}{4} x + 16 + x^{2}) + C

求解步骤

\int \frac{- 6}{x ^{2} + 16} d x

提出常数:

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

= - 6 \cdot \int \frac{1}{x ^{2} + 16} d x

使用三角换元法

= - 6 \cdot \int sec (u) d u

使用积分常见定则:

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

= - 6 ln ∣ tan (u) + sec (u) ∣

u = arctan (\frac{1}{4} x)

代回

= - 6 ln tan (arctan (\frac{1}{4} x)) + sec (arctan (\frac{1}{4} x))

化简

- 6 ln tan (arctan (\frac{1}{4} x)) + sec (arctan (\frac{1}{4} x)) : - 6 ln (\frac{1}{4} x + 16 + x^{2})

= - 6 ln (\frac{1}{4} x + 16 + x^{2})

解答补常数

= - 6 ln (\frac{1}{4} x + 16 + x^{2}) + C

x \to 1 + lim (\frac{2}{1 - x})

\frac{d}{d x} (a - b)

d er i v a t i v e g (t) = t^{2} - \frac{2}{t ^{3}}

\int t^{2} e^{- 20 t} d t

\int_{t}^{1} ln (x) d x