integral of (x+4)/(sqrt(x+1))

解答

\int \frac{x + 4}{x + 1} d x

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

求解步骤

\int \frac{x + 4}{x + 1} d x

使用换元积分法

= \int 2 (u^{2} + 3) 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} + 3 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 3 d u)

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

\int 3 d u = 3 u

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

u = x + 1

代回

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

化简

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

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

解答补常数

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

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

\frac{d}{d x} (e^{- x} cos (x))

\int_{0}^{4} (2 x + 5) d x

d er i v a t i v e x^{2} f (x)

in v erse l a pl a ce \frac{1}{( s ^{2} + 4 ) ^{2} \cdot s}