integral of-7/(sqrt(x^2+16))

解答

\int - \frac{7}{x ^{2} + 16} d x

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

求解步骤

\int - \frac{7}{x ^{2} + 16} d x

提出常数:

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

= - 7 \cdot \int \frac{1}{x ^{2} + 16} d x

使用三角换元法

= - 7 \cdot \int sec (u) d u

使用积分常见定则:

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

= - 7 ln ∣ tan (u) + sec (u) ∣

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

代回

= - 7 ln tan (arctan (\frac{1}{4} x)) + sec (arctan (\frac{1}{4} x))

化简

- 7 ln tan (arctan (\frac{1}{4} x)) + sec (arctan (\frac{1}{4} x)) : - 7 ln (\frac{1}{4} x + 16 + x^{2})

= - 7 ln (\frac{1}{4} x + 16 + x^{2})

解答补常数

= - 7 ln (\frac{1}{4} x + 16 + x^{2}) + C

t an g e n t f (x) = \frac{1}{2 x}, at x = 64

\frac{d}{d x} (cos (x + π))

\frac{d}{d x} (x^{2} + 4 x - 5)

\int sec^{3} (4 x) tan (4 x) d x

\int \frac{1}{5 x} d x