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人気のある微分積分 >

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

解

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

解

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

解答ステップ

\int \frac{x ^{2}}{2 x - x ^{2}} d x

平方完成する

2 x - x^{2} : - (x - 1)^{2} + 1

= \int \frac{x ^{2}}{- ( x - 1 ) ^{2} + 1} d x

u置換積分法を適用する

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

拡張

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

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

総和規則を適用する:

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

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

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

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

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

代用を戻す

u = x - 1

= \frac{1}{2} (arcsin (x - 1) - \frac{1}{2} sin (2 arcsin (x - 1))) - 2 - (x - 1)^{2} + 1 + arcsin (x - 1)

簡素化

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

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

定数を解答に追加する

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

グラフ

人気の例

integral of 60x

\int 60 x d x

integral of 1/(sqrt(x(3x+5)))

\int \frac{1}{x ( 3 x + 5 )} d x

integral of 2tan(2x)

\int 2 tan (2 x) d x

integral of ((x^2cos(x)-2xsin(x)))/(x^4)

\int \frac{( x ^{2} cos ( x ) - 2 x sin ( x ) )}{x ^{4}} d x

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