integral of 1/(sqrt(x^2+3))

解答

\int \frac{1}{x ^{2} + 3} d x

ln \frac{1}{3} x + \frac{1}{3} (3 + x^{2}) + C

求解步骤

\int \frac{1}{x ^{2} + 3} 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}{3} x)

代回

= ln tan (arctan (\frac{1}{3} x)) + sec (arctan (\frac{1}{3} x))

化简

ln tan (arctan (\frac{1}{3} x)) + sec (arctan (\frac{1}{3} x)) : ln \frac{1}{3} x + \frac{1}{3} (3 + x^{2})

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

解答补常数

= ln \frac{1}{3} x + \frac{1}{3} (3 + x^{2}) + C

θ \to 0 lim (θ cos (\frac{3}{θ}))

\int \frac{( x - 3 ) ^{3}}{x} d x

\int 8 x d x

\int \frac{9}{3 + 9 x} d x

\int \frac{1}{1 - 4 y} d y