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人気のある微分積分 >

integral of x/(sqrt(x^2+6x+12))

解

\int \frac{x}{x ^{2} + 6 x + 12} d x

解

x^{2} + 6 x + 12 - 3 ln \frac{1}{3} (x + 3) + \frac{1}{3} (x^{2} + 6 x + 12) + C

解答ステップ

\int \frac{x}{x ^{2} + 6 x + 12} d x

平方完成する

x^{2} + 6 x + 12 : (x + 3)^{2} + 3

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

u置換積分法を適用する

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

拡張

\frac{u - 3}{u ^{2} + 3} : \frac{u}{u ^{2} + 3} - \frac{3}{u ^{2} + 3}

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

\int \frac{u}{u ^{2} + 3} d u = u^{2} + 3

\int \frac{3}{u ^{2} + 3} d u = 3 ln \frac{1}{3} u + \frac{1}{3} (3 + u^{2})

= u^{2} + 3 - 3 ln \frac{1}{3} u + \frac{1}{3} (3 + u^{2})

代用を戻す

u = x + 3

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

簡素化

(x + 3)^{2} + 3 - 3 ln \frac{1}{3} (x + 3) + \frac{1}{3} (3 + (x + 3)^{2}) : x^{2} + 6 x + 12 - 3 ln \frac{1}{3} (x + 3) + \frac{1}{3} (x^{2} + 6 x + 12)

= x^{2} + 6 x + 12 - 3 ln \frac{1}{3} (x + 3) + \frac{1}{3} (x^{2} + 6 x + 12)

定数を解答に追加する

= x^{2} + 6 x + 12 - 3 ln \frac{1}{3} (x + 3) + \frac{1}{3} (x^{2} + 6 x + 12) + C

グラフ

人気の例

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