integral of 1/(sqrt(x^2+625))

解答

\int \frac{1}{x ^{2} + 625} d x

ln (\frac{1}{25} x + 625 + x^{2}) + C

求解步骤

\int \frac{1}{x ^{2} + 625} 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}{25} x)

代回

= ln tan (arctan (\frac{1}{25} x)) + sec (arctan (\frac{1}{25} x))

化简

ln tan (arctan (\frac{1}{25} x)) + sec (arctan (\frac{1}{25} x)) : ln (\frac{1}{25} x + 625 + x^{2})

= ln (\frac{1}{25} x + 625 + x^{2})

解答补常数

= ln (\frac{1}{25} x + 625 + x^{2}) + C

x \to 0 lim (\frac{∣ x ∣}{x ^{3} - 5})

t an g e n t y = 8 x^{2} - 5 x^{11}

\int (x - 2) e^{x + 3} d x

\frac{\partial}{\partial x} (4 x^{2} - y^{2} + 2 y)

a re a y = \frac{x}{3}, y = x - 2