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受欢迎的微积分 >

integral of (x*e^{arcsin(x)})/(sqrt(1-x^2))

解答

\int \frac{x \cdot e ^{a r c s i n (x)}}{1 - x ^{2}} d x

解答

- \frac{1}{2} e^{a r c s i n (x)} 1 - x^{2} + \frac{1}{2} e^{a r c s i n (x)} x + C

求解步骤

\int \frac{x e ^{a r c s i n (x)}}{1 - x ^{2}} d x

使用换元积分法

= \int e^{u} sin (u) d u

使用分布积分法

= - e^{u} cos (u) - \int - e^{u} cos (u) d u

提出常数:

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

= - e^{u} cos (u) - (- \int e^{u} cos (u) d u)

使用分布积分法

= - e^{u} cos (u) - (- (e^{u} sin (u) - \int e^{u} sin (u) d u))

因此

\int e^{u} sin (u) d u = - e^{u} cos (u) - (- (e^{u} sin (u) - \int e^{u} sin (u) d u))

整理

\int e^{u} sin (u) d u

= - \frac{e ^{u} cos ( u )}{2} + \frac{e ^{u} sin ( u )}{2}

u = arcsin (x)

代回

= - \frac{e ^{a r c s i n (x)} cos ( arcsin ( x ) )}{2} + \frac{e ^{a r c s i n (x)} sin ( arcsin ( x ) )}{2}

化简

- \frac{e ^{a r c s i n (x)} cos ( arcsin ( x ) )}{2} + \frac{e ^{a r c s i n (x)} sin ( arcsin ( x ) )}{2} : - \frac{1}{2} e^{a r c s i n (x)} 1 - x^{2} + \frac{1}{2} e^{a r c s i n (x)} x

= - \frac{1}{2} e^{a r c s i n (x)} 1 - x^{2} + \frac{1}{2} e^{a r c s i n (x)} x

解答补常数

= - \frac{1}{2} e^{a r c s i n (x)} 1 - x^{2} + \frac{1}{2} e^{a r c s i n (x)} x + C

作图

流行的例子

integral of 1/x-3/(16x^2+1)

\int \frac{1}{x} - \frac{3}{16 x ^{2} + 1} d x

integral of e^{2x}e^{e^{2x}}

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

integral of (x+10)sqrt(20x+x^2)

\int (x + 10) 20 x + x^{2} d x

integral of (x^2-x-21)/(2x^3-x^2+8x-4)

\int \frac{x ^{2} - x - 21}{2 x ^{3} - x ^{2} + 8 x - 4} d x

integral of 1/(4x^{3/4)}

\int \frac{1}{4 x ^{\frac{3}{4}}} d x