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人気のある微分積分 >

integral of (5x-15)/((x^2+2x+4)^2)

解

\int \frac{5 x - 15}{( x ^{2} + 2 x + 4 ) ^{2}} d x

解

- \frac{5}{2 ( x ^{2} + 2 x + 4 )} - \frac{10}{3 3} (arctan (\frac{1}{3} (x + 1)) + \frac{1}{2} sin (2 arctan (\frac{1}{3} (x + 1)))) + C

解答ステップ

\int \frac{5 x - 15}{( x ^{2} + 2 x + 4 ) ^{2}} d x

平方完成する

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

= \int \frac{5 x - 15}{( ( x + 1 ) ^{2} + 3 ) ^{2}} d x

u置換積分法を適用する

= \int \frac{5 u - 20}{( u ^{2} + 3 ) ^{2}} d u

拡張

\frac{5 u - 20}{( u ^{2} + 3 ) ^{2}} : \frac{5 u}{( u ^{2} + 3 ) ^{2}} - \frac{20}{( u ^{2} + 3 ) ^{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{5 u}{( u ^{2} + 3 ) ^{2}} d u - \int \frac{20}{( u ^{2} + 3 ) ^{2}} d u

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

\int \frac{20}{( u ^{2} + 3 ) ^{2}} d u = \frac{10}{3 3} (arctan (\frac{u}{3}) + \frac{1}{2} sin (2 arctan (\frac{u}{3})))

= - \frac{5}{2 ( u ^{2} + 3 )} - \frac{10}{3 3} (arctan (\frac{u}{3}) + \frac{1}{2} sin (2 arctan (\frac{u}{3})))

代用を戻す

u = x + 1

= - \frac{5}{2 ( ( x + 1 ) ^{2} + 3 )} - \frac{10}{3 3} (arctan (\frac{x + 1}{3}) + \frac{1}{2} sin (2 arctan (\frac{x + 1}{3})))

簡素化

- \frac{5}{2 ( ( x + 1 ) ^{2} + 3 )} - \frac{10}{3 3} (arctan (\frac{x + 1}{3}) + \frac{1}{2} sin (2 arctan (\frac{x + 1}{3}))) : - \frac{5}{2 ( x ^{2} + 2 x + 4 )} - \frac{10}{3 3} (arctan (\frac{1}{3} (x + 1)) + \frac{1}{2} sin (2 arctan (\frac{1}{3} (x + 1))))

= - \frac{5}{2 ( x ^{2} + 2 x + 4 )} - \frac{10}{3 3} (arctan (\frac{1}{3} (x + 1)) + \frac{1}{2} sin (2 arctan (\frac{1}{3} (x + 1))))

定数を解答に追加する

= - \frac{5}{2 ( x ^{2} + 2 x + 4 )} - \frac{10}{3 3} (arctan (\frac{1}{3} (x + 1)) + \frac{1}{2} sin (2 arctan (\frac{1}{3} (x + 1)))) + C

グラフ

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