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人気のある微分積分 >

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

解

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

解

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

解答ステップ

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

平方完成する

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

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

u置換積分法を適用する

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

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{2 u - 1}{4 u ^{2} + 2} d u

拡張

\frac{2 u - 1}{4 u ^{2} + 2} : \frac{2 u}{4 u ^{2} + 2} - \frac{1}{4 u ^{2} + 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

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

\int \frac{2 u}{4 u ^{2} + 2} d u = \frac{1}{4} ln 4 u^{2} + 2

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

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

代用を戻す

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

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

簡素化

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

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

定数を解答に追加する

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

グラフ

人気の例

y^'=(2x+1)/(x^2-3x+2),y(3)=0

y^{^{'}} = \frac{2 x + 1}{x ^{2} - 3 x + 2}, y (3) = 0

x(dy)/(dx)=1+3y+y^2

x \frac{d y}{d x} = 1 + 3 y + y^{2}

derivative of xcos(x+2tan(x))

\frac{d}{d x} (x cos (x) + 2 tan (x))

tangent-x^2+7x+2,\at x=3

t an g e n t - x^{2} + 7 x + 2, at x = 3

tangent y=2x^3-x^2+1,(1,2)

t an g e n t y = 2 x^{3} - x^{2} + 1, (1, 2)