integral of 1/(sqrt(\sqrt{x)-1)}

解答

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

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

求解步骤

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

使用换元积分法

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

提出常数:

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

= 4 \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

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

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

\int 1 d u = u

= 4 (\frac{u ^{3}}{3} + u)

u = x - 1

代回

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

化简

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

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

解答补常数

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

\int - 3 (x + 2)^{- 2} d x

\int (\frac{x ^{2} - 3 x + 5}{x}) d x

\int (x + 2)^{2} \cdot ln (5 x) d x

\int tan (t) cot (t) d t

\int \frac{x ^{3}}{x ^{2} + 3} d x