integral of x/((sqrt(x+1)))

解答

\int \frac{x}{( x + 1 )} d x

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

求解步骤

\int \frac{x}{x + 1} d x

使用换元积分法

= \int 2 (u^{2} - 1) d u

提出常数:

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

= 2 \cdot \int u^{2} - 1 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

= 2 (\int u^{2} d u - \int 1 d u)

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

\int 1 d u = u

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

u = x + 1

代回

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

化简

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

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

解答补常数

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

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