integral of 1/(sqrt(x^2+100))

解答

\int \frac{1}{x ^{2} + 100} d x

ln (\frac{1}{10} x + 100 + x^{2}) + C

求解步骤

\int \frac{1}{x ^{2} + 100} 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}{10} x)

代回

= ln tan (arctan (\frac{1}{10} x)) + sec (arctan (\frac{1}{10} x))

化简

ln tan (arctan (\frac{1}{10} x)) + sec (arctan (\frac{1}{10} x)) : ln (\frac{1}{10} x + 100 + x^{2})

= ln (\frac{1}{10} x + 100 + x^{2})

解答补常数

= ln (\frac{1}{10} x + 100 + x^{2}) + C

t y^{^{'}} = 2 y + 2 t^{3} sin (3 t), y (\frac{π}{3}) = 0

\int \frac{e ^{5 t}}{t} d t

x \to 0 - lim (\frac{x + 3}{x})

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

d er i v a t i v e 3 t^{2} - 12