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人気のある微分積分 >

integral of (x+2)/((x^2+2x+10)^{5/2)}

解

\int \frac{x + 2}{( x ^{2} + 2 x + 10 ) ^{\frac{5}{2}}} d x

解

- \frac{1}{3 ( x ^{2} + 2 x + 10 ) ^{\frac{3}{2}}} + \frac{2 ( x + 1 ) ^{3} + 27 ( x + 1 )}{243 ( x ^{2} + 2 x + 10 ) ^{\frac{3}{2}}} + C

解答ステップ

\int \frac{x + 2}{( x ^{2} + 2 x + 10 ) ^{\frac{5}{2}}} d x

平方完成する

x^{2} + 2 x + 10 : (x + 1)^{2} + 9

= \int \frac{x + 2}{( ( x + 1 ) ^{2} + 9 ) ^{\frac{5}{2}}} d x

u置換積分法を適用する

= \int \frac{u + 1}{( u ^{2} + 9 ) ^{\frac{5}{2}}} d u

拡張

\frac{u + 1}{( u ^{2} + 9 ) ^{\frac{5}{2}}} : \frac{u}{( u ^{2} + 9 ) ^{\frac{5}{2}}} + \frac{1}{( u ^{2} + 9 ) ^{\frac{5}{2}}}

= \int \frac{u}{( u ^{2} + 9 ) ^{\frac{5}{2}}} + \frac{1}{( u ^{2} + 9 ) ^{\frac{5}{2}}} 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}{( u ^{2} + 9 ) ^{\frac{5}{2}}} d u + \int \frac{1}{( u ^{2} + 9 ) ^{\frac{5}{2}}} d u

\int \frac{u}{( u ^{2} + 9 ) ^{\frac{5}{2}}} d u = - \frac{1}{3 ( u ^{2} + 9 ) ^{\frac{3}{2}}}

\int \frac{1}{( u ^{2} + 9 ) ^{\frac{5}{2}}} d u = \frac{2 u ^{3} + 27 u}{243 ( u ^{2} + 9 ) ^{\frac{3}{2}}}

= - \frac{1}{3 ( u ^{2} + 9 ) ^{\frac{3}{2}}} + \frac{2 u ^{3} + 27 u}{243 ( u ^{2} + 9 ) ^{\frac{3}{2}}}

代用を戻す

u = x + 1

= - \frac{1}{3 ( ( x + 1 ) ^{2} + 9 ) ^{\frac{3}{2}}} + \frac{2 ( x + 1 ) ^{3} + 27 ( x + 1 )}{243 ( ( x + 1 ) ^{2} + 9 ) ^{\frac{3}{2}}}

簡素化

- \frac{1}{3 ( ( x + 1 ) ^{2} + 9 ) ^{\frac{3}{2}}} + \frac{2 ( x + 1 ) ^{3} + 27 ( x + 1 )}{243 ( ( x + 1 ) ^{2} + 9 ) ^{\frac{3}{2}}} : - \frac{1}{3 ( x ^{2} + 2 x + 10 ) ^{\frac{3}{2}}} + \frac{2 ( x + 1 ) ^{3} + 27 ( x + 1 )}{243 ( x ^{2} + 2 x + 10 ) ^{\frac{3}{2}}}

= - \frac{1}{3 ( x ^{2} + 2 x + 10 ) ^{\frac{3}{2}}} + \frac{2 ( x + 1 ) ^{3} + 27 ( x + 1 )}{243 ( x ^{2} + 2 x + 10 ) ^{\frac{3}{2}}}

定数を解答に追加する

= - \frac{1}{3 ( x ^{2} + 2 x + 10 ) ^{\frac{3}{2}}} + \frac{2 ( x + 1 ) ^{3} + 27 ( x + 1 )}{243 ( x ^{2} + 2 x + 10 ) ^{\frac{3}{2}}} + C

グラフ

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