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人気のある微分積分 >

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

解

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

解

\frac{1}{x - 2} - \frac{1}{2} ln ∣ x - 1 ∣ + \frac{1}{2} ln ∣ x - 3 ∣ + C

解答ステップ

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

平方完成する

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

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

u置換積分法を適用する

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

以下の部分分数を得る:

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

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

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

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

\int \frac{1}{2 ( u - 1 )} d u = \frac{1}{2} ln ∣ u - 1 ∣

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

代用を戻す

u = x - 2

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

簡素化

- (- \frac{1}{x - 2}) - \frac{1}{2} ln ∣ x - 2 + 1 ∣ + \frac{1}{2} ln ∣ x - 2 - 1 ∣ : \frac{1}{x - 2} - \frac{1}{2} ln ∣ x - 1 ∣ + \frac{1}{2} ln ∣ x - 3 ∣

= \frac{1}{x - 2} - \frac{1}{2} ln ∣ x - 1 ∣ + \frac{1}{2} ln ∣ x - 3 ∣

定数を解答に追加する

= \frac{1}{x - 2} - \frac{1}{2} ln ∣ x - 1 ∣ + \frac{1}{2} ln ∣ x - 3 ∣ + C

グラフ

人気の例

derivative f(x)=x^3+4x^2+3x

d er i v a t i v e f (x) = x^{3} + 4 x^{2} + 3 x

integral of (9-x^2)/x

\int \frac{9 - x ^{2}}{x} d x

derivative of (x-1(x^2+2)^3)

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

(\partial)/(\partial x)(ln(2+x^2y^2))

\frac{\partial}{\partial x} (ln (2 + x^{2} y^{2}))

integral of (6x^3)/(sqrt(x^2+9))

\int \frac{6 x ^{3}}{x ^{2} + 9} d x