integral of x/(sqrt(a+bx))

解答

\int \frac{x}{a + b x} d x

\frac{2}{b ^{2}} (\frac{1}{3} (a + b x)^{\frac{3}{2}} - a a + b x) + C

求解步骤

\int \frac{x}{a + b x} d x

使用换元积分法

= \int \frac{2 ( u ^{2} - a )}{b ^{2}} d u

提出常数:

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

= \frac{2}{b ^{2}} \cdot \int u^{2} - a d u

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{2}{b ^{2}} (\int u^{2} d u - \int a d u)

\int u^{2} d u = \frac{u ^{3}}{3}

\int a d u = a u

= \frac{2}{b ^{2}} (\frac{u ^{3}}{3} - a u)

u = a + b x

代回

= \frac{2}{b ^{2}} (\frac{( a + b x ) ^{3}}{3} - a a + b x)

化简

\frac{2}{b ^{2}} (\frac{( a + b x ) ^{3}}{3} - a a + b x) : \frac{2}{b ^{2}} (\frac{1}{3} (a + b x)^{\frac{3}{2}} - a a + b x)

= \frac{2}{b ^{2}} (\frac{1}{3} (a + b x)^{\frac{3}{2}} - a a + b x)

解答补常数

= \frac{2}{b ^{2}} (\frac{1}{3} (a + b x)^{\frac{3}{2}} - a a + b x) + C

im pl i c i t \frac{d y}{d x}, x^{2} y = 1

\int \frac{2 x}{λ ^{2}} d x

x \to 0 lim (\frac{e ^{4 x} - 1}{2 x ^{2} + 2 x})

\int sin (x) sin (n) x d x

n = 1 \sum \infty \frac{4}{n ( n + 6 )}