integral of (u^2)/(sqrt(1-u^2))

解答

\int \frac{u ^{2}}{1 - u ^{2}} d u

\frac{1}{2} (arcsin (u) - \frac{1}{2} sin (2 arcsin (u))) + C

求解步骤

\int \frac{u ^{2}}{1 - u ^{2}} d u

使用三角换元法

= \int sin^{2} (v) d v

使用三角恒等式改写

= \int \frac{1 - cos ( 2 v )}{2} d v

提出常数:

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

= \frac{1}{2} \cdot \int 1 - cos (2 v) d v

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{1}{2} (\int 1 d v - \int cos (2 v) d v)

\int 1 d v = v

\int cos (2 v) d v = \frac{1}{2} sin (2 v)

= \frac{1}{2} (v - \frac{1}{2} sin (2 v))

v = arcsin (u)

代回

= \frac{1}{2} (arcsin (u) - \frac{1}{2} sin (2 arcsin (u)))

解答补常数

= \frac{1}{2} (arcsin (u) - \frac{1}{2} sin (2 arcsin (u))) + C

\int \frac{16}{1 + 4 x ^{2}} d x

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

\int \frac{6 x ^{2}}{x ^{3} - 4} d x

\int (2 - x) e^{x} d x

\int \frac{- x ^{2} + 5 x - 4}{( 2 - x ) ^{3}} d x