integral of x/(sqrt(x^2+4))

解答

\int \frac{x}{x ^{2} + 4} d x

x^{2} + 4 + C

求解步骤

\int \frac{x}{x ^{2} + 4} d x

使用换元积分法

= \int \frac{1}{2 u} d u

提出常数:

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

= \frac{1}{2} \cdot \int \frac{1}{u} d u

使用幂法则

= \frac{1}{2} \cdot 2 u^{\frac{1}{2}}

u = x^{2} + 4

代回

= \frac{1}{2} \cdot 2 (x^{2} + 4)^{\frac{1}{2}}

化简

\frac{1}{2} \cdot 2 (x^{2} + 4)^{\frac{1}{2}} : x^{2} + 4

= x^{2} + 4

解答补常数

= x^{2} + 4 + C

x \to 2 lim (- \frac{1}{( x - 2 ) ^{2}})

\frac{d}{d x} (tan (\frac{1}{2} x))

\int sin (x + 3) d x

\frac{\partial}{\partial x} (3 x^{2} + 2 y^{2} - 4 x y + 5)

\frac{d}{d t} (\frac{t}{1 + ( 1.5574 + 5.43 t ^{2} ) ^{2}})