integral of 1/(sqrt(x^2+169))

解答

\int \frac{1}{x ^{2} + 169} d x

ln (\frac{1}{13} x + 169 + x^{2}) + C

求解步骤

\int \frac{1}{x ^{2} + 169} d x

使用三角换元法

= \int sec (u) d u

使用积分常见定则:

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

= ln ∣ tan (u) + sec (u) ∣

u = arctan (\frac{1}{13} x)

代回

= ln tan (arctan (\frac{1}{13} x)) + sec (arctan (\frac{1}{13} x))

化简

ln tan (arctan (\frac{1}{13} x)) + sec (arctan (\frac{1}{13} x)) : ln (\frac{1}{13} x + 169 + x^{2})

= ln (\frac{1}{13} x + 169 + x^{2})

解答补常数

= ln (\frac{1}{13} x + 169 + x^{2}) + C

(2 x + 1) (\frac{d y}{d x}) + y = (2 x + 1)^{\frac{3}{2}}

y^{^{''}} + 3 y = - 48 x^{2} e^{3 x}

f (x) = arcsinh (x)

\int - 510 x^{2} + 3414 x + 1998 d x

d er i v a t i v e f (x) = (9 - x^{3})^{- 2}