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人気のある微分積分 >

integral of x/((x^2+2x+2)^2)

解

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

解

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

解答ステップ

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

平方完成する

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

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

u置換積分法を適用する

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

拡張

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

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

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

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

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

代用を戻す

u = x + 1

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

簡素化

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

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

定数を解答に追加する

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

グラフ

人気の例

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\int_{0}^{6.25} - x^{2} + 5 x d x

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\int 15 x^{\frac{1}{4}} d x

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