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受欢迎的微积分 >

integral of 4/(sqrt(x+1)+\sqrt{x)}

解答

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

解答

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

求解步骤

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

提出常数:

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

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

使用换元积分法

= 4 \cdot \int \frac{( u + 1 ) ( u - 1 ) ( u ^{2} + 1 )}{2 u ^{4}} d u

提出常数:

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

= 4 \cdot \frac{1}{2} \cdot \int \frac{( u + 1 ) ( u - 1 ) ( u ^{2} + 1 )}{u ^{4}} d u

乘开

\frac{( u + 1 ) ( u - 1 ) ( u ^{2} + 1 )}{u ^{4}} : 1 - \frac{1}{u ^{4}}

= 4 \cdot \frac{1}{2} \cdot \int 1 - \frac{1}{u ^{4}} 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

= 4 \cdot \frac{1}{2} (\int 1 d u - \int \frac{1}{u ^{4}} d u)

\int 1 d u = u

\int \frac{1}{u ^{4}} d u = - \frac{1}{3 u ^{3}}

= 4 \cdot \frac{1}{2} (u - (- \frac{1}{3 u ^{3}}))

u = x + 1 + x

代回

= 4 \cdot \frac{1}{2} (x + 1 + x - (- \frac{1}{3 ( x + 1 + x ) ^{3}}))

化简

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

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

解答补常数

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

作图

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