integral of (x^2)/(sqrt((x-3)(5-x)))

解答

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

\frac{3}{2} arcsin (x - 4) - \frac{1}{4} sin (2 arcsin (x - 4)) - 8 - x^{2} + 8 x - 15 + 15 arcsin (x - 4) + C

求解步骤

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

使用换元积分法

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

乘开

\frac{( u + 3 ) ^{2}}{u ( - u + 2 )} : \frac{u ^{2}}{- u ^{2} + 2 u} + \frac{6 u}{- u ^{2} + 2 u} + \frac{9}{- u ^{2} + 2 u}

= \int \frac{u ^{2}}{- u ^{2} + 2 u} + \frac{6 u}{- u ^{2} + 2 u} + \frac{9}{- u ^{2} + 2 u} 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

= \int \frac{u ^{2}}{- u ^{2} + 2 u} d u + \int \frac{6 u}{- u ^{2} + 2 u} d u + \int \frac{9}{- u ^{2} + 2 u} d u

\int \frac{u ^{2}}{- u ^{2} + 2 u} d u = \frac{3}{2} arcsin (u - 1) - \frac{1}{4} sin (2 arcsin (u - 1)) - 2 - u^{2} + 2 u

\int \frac{6 u}{- u ^{2} + 2 u} d u = 6 (- - u^{2} + 2 u + arcsin (u - 1))

\int \frac{9}{- u ^{2} + 2 u} d u = 9 arcsin (u - 1)

= \frac{3}{2} arcsin (u - 1) - \frac{1}{4} sin (2 arcsin (u - 1)) - 2 - u^{2} + 2 u + 6 (- - u^{2} + 2 u + arcsin (u - 1)) + 9 arcsin (u - 1)

u = x - 3

代回

= \frac{3}{2} arcsin (x - 3 - 1) - \frac{1}{4} sin (2 arcsin (x - 3 - 1)) - 2 - (x - 3)^{2} + 2 (x - 3) + 6 (- - (x - 3)^{2} + 2 (x - 3) + arcsin (x - 3 - 1)) + 9 arcsin (x - 3 - 1)

化简

\frac{3}{2} arcsin (x - 3 - 1) - \frac{1}{4} sin (2 arcsin (x - 3 - 1)) - 2 - (x - 3)^{2} + 2 (x - 3) + 6 (- - (x - 3)^{2} + 2 (x - 3) + arcsin (x - 3 - 1)) + 9 arcsin (x - 3 - 1) : \frac{3}{2} arcsin (x - 4) - \frac{1}{4} sin (2 arcsin (x - 4)) - 8 - x^{2} + 8 x - 15 + 15 arcsin (x - 4)

= \frac{3}{2} arcsin (x - 4) - \frac{1}{4} sin (2 arcsin (x - 4)) - 8 - x^{2} + 8 x - 15 + 15 arcsin (x - 4)

解答补常数

= \frac{3}{2} arcsin (x - 4) - \frac{1}{4} sin (2 arcsin (x - 4)) - 8 - x^{2} + 8 x - 15 + 15 arcsin (x - 4) + C

\int \frac{5}{( x - 7 ) ( x - 2 )} d x

\int \frac{sin ( e ^{- x} )}{2 e ^{3 x}} d x

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

\int (3 x - \frac{1}{x}) d x

\int 3 e^{5 x - 2} cos (7 - 4 x) d x