integral of x/(sqrt(ax+b))

解答

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

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

求解步骤

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

使用换元积分法

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

提出常数:

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

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

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

\int b d u = b u

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

u = a x + b

代回

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

化简

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

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

解答补常数

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

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