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人気のある微分積分 >

integral of ((2x-3))/((x^2+2x-3)^{3/2)}

解

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

解

- \frac{2}{( x ^{2} + 2 x - 3 ) ^{\frac{1}{2}}} + \frac{5 ( x + 1 )}{4 ( x ^{2} + 2 x - 3 ) ^{\frac{1}{2}}} + C

解答ステップ

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

平方完成する

x^{2} + 2 x - 3 : (x + 1)^{2} - 4

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

u置換積分法を適用する

= \int \frac{2 u - 5}{( u ^{2} - 4 ) ^{\frac{3}{2}}} d u

拡張

\frac{2 u - 5}{( u ^{2} - 4 ) ^{\frac{3}{2}}} : \frac{2 u}{( u ^{2} - 4 ) ^{\frac{3}{2}}} - \frac{5}{( u ^{2} - 4 ) ^{\frac{3}{2}}}

= \int \frac{2 u}{( u ^{2} - 4 ) ^{\frac{3}{2}}} - \frac{5}{( u ^{2} - 4 ) ^{\frac{3}{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{2 u}{( u ^{2} - 4 ) ^{\frac{3}{2}}} d u - \int \frac{5}{( u ^{2} - 4 ) ^{\frac{3}{2}}} d u

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

\int \frac{5}{( u ^{2} - 4 ) ^{\frac{3}{2}}} d u = - \frac{5 u}{4 ( u ^{2} - 4 ) ^{\frac{1}{2}}}

= - \frac{2}{( u ^{2} - 4 ) ^{\frac{1}{2}}} - (- \frac{5 u}{4 ( u ^{2} - 4 ) ^{\frac{1}{2}}})

代用を戻す

u = x + 1

= - \frac{2}{( ( x + 1 ) ^{2} - 4 ) ^{\frac{1}{2}}} - - \frac{5 ( x + 1 )}{4 ( ( x + 1 ) ^{2} - 4 ) ^{\frac{1}{2}}}

簡素化

- \frac{2}{( ( x + 1 ) ^{2} - 4 ) ^{\frac{1}{2}}} - - \frac{5 ( x + 1 )}{4 ( ( x + 1 ) ^{2} - 4 ) ^{\frac{1}{2}}} : - \frac{2}{( x ^{2} + 2 x - 3 ) ^{\frac{1}{2}}} + \frac{5 ( x + 1 )}{4 ( x ^{2} + 2 x - 3 ) ^{\frac{1}{2}}}

= - \frac{2}{( x ^{2} + 2 x - 3 ) ^{\frac{1}{2}}} + \frac{5 ( x + 1 )}{4 ( x ^{2} + 2 x - 3 ) ^{\frac{1}{2}}}

定数を解答に追加する

= - \frac{2}{( x ^{2} + 2 x - 3 ) ^{\frac{1}{2}}} + \frac{5 ( x + 1 )}{4 ( x ^{2} + 2 x - 3 ) ^{\frac{1}{2}}} + C

グラフ

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