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人気のある微分積分 >

integral of (3x-1)/(4x^2-4x+12)

解

\int \frac{3 x - 1}{4 x ^{2} - 4 x + 12} d x

解

3 (\frac{1}{8} ln 4 x^{2} - 4 x + 12 + \frac{1}{4 11} arctan (\frac{2 x - 1}{11})) - \frac{1}{2 11} arctan (\frac{2 x - 1}{11}) + C

解答ステップ

\int \frac{3 x - 1}{4 x ^{2} - 4 x + 12} d x

平方完成する

4 x^{2} - 4 x + 12 : 4 (x - \frac{1}{2})^{2} + 11

= \int \frac{3 x - 1}{4 ( x - \frac{1}{2} ) ^{2} + 11} d x

u置換積分法を適用する

= \int \frac{3 ( u + \frac{1}{2} ) - 1}{4 u ^{2} + 11} d u

拡張

\frac{3 ( u + \frac{1}{2} ) - 1}{4 u ^{2} + 11} : \frac{3 ( u + \frac{1}{2} )}{4 u ^{2} + 11} - \frac{1}{4 u ^{2} + 11}

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{3 ( u + \frac{1}{2} )}{4 u ^{2} + 11} d u - \int \frac{1}{4 u ^{2} + 11} d u

\int \frac{3 ( u + \frac{1}{2} )}{4 u ^{2} + 11} d u = 3 (\frac{1}{8} ln 4 u^{2} + 11 + \frac{1}{4 11} arctan (\frac{2}{11} u))

\int \frac{1}{4 u ^{2} + 11} d u = \frac{1}{2 11} arctan (\frac{2}{11} u)

= 3 (\frac{1}{8} ln 4 u^{2} + 11 + \frac{1}{4 11} arctan (\frac{2}{11} u)) - \frac{1}{2 11} arctan (\frac{2}{11} u)

代用を戻す

u = x - \frac{1}{2}

= 3 (\frac{1}{8} ln 4 (x - \frac{1}{2})^{2} + 11 + \frac{1}{4 11} arctan (\frac{2}{11} (x - \frac{1}{2}))) - \frac{1}{2 11} arctan (\frac{2}{11} (x - \frac{1}{2}))

簡素化

3 (\frac{1}{8} ln 4 (x - \frac{1}{2})^{2} + 11 + \frac{1}{4 11} arctan (\frac{2}{11} (x - \frac{1}{2}))) - \frac{1}{2 11} arctan (\frac{2}{11} (x - \frac{1}{2})) : 3 (\frac{1}{8} ln 4 x^{2} - 4 x + 12 + \frac{1}{4 11} arctan (\frac{2 x - 1}{11})) - \frac{1}{2 11} arctan (\frac{2 x - 1}{11})

= 3 (\frac{1}{8} ln 4 x^{2} - 4 x + 12 + \frac{1}{4 11} arctan (\frac{2 x - 1}{11})) - \frac{1}{2 11} arctan (\frac{2 x - 1}{11})

定数を解答に追加する

= 3 (\frac{1}{8} ln 4 x^{2} - 4 x + 12 + \frac{1}{4 11} arctan (\frac{2 x - 1}{11})) - \frac{1}{2 11} arctan (\frac{2 x - 1}{11}) + C

グラフ

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