integral of (x^6)/((x-2)^2(1-x)^5)

解

\int \frac{x ^{6}}{( x - 2 ) ^{2} ( 1 - x ) ^{5}} d x

128 ln ∣ x - 2 ∣ + \frac{64}{x - 2} - 129 ln ∣ x - 1 ∣ + \frac{72}{x - 1} + \frac{15}{( x - 1 ) ^{2}} + \frac{8}{3 ( x - 1 ) ^{3}} + \frac{1}{4 ( x - 1 ) ^{4}} + C

解答ステップ

\int \frac{x ^{6}}{( x - 2 ) ^{2} ( 1 - x ) ^{5}} d x

以下の部分分数を得る:

\frac{x ^{6}}{( x - 2 ) ^{2} ( 1 - x ) ^{5}} : \frac{128}{x - 2} - \frac{64}{( x - 2 ) ^{2}} - \frac{129}{x - 1} - \frac{72}{( x - 1 ) ^{2}} - \frac{30}{( x - 1 ) ^{3}} - \frac{8}{( x - 1 ) ^{4}} - \frac{1}{( x - 1 ) ^{5}}

= \int \frac{128}{x - 2} - \frac{64}{( x - 2 ) ^{2}} - \frac{129}{x - 1} - \frac{72}{( x - 1 ) ^{2}} - \frac{30}{( x - 1 ) ^{3}} - \frac{8}{( x - 1 ) ^{4}} - \frac{1}{( x - 1 ) ^{5}} d x

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{128}{x - 2} d x - \int \frac{64}{( x - 2 ) ^{2}} d x - \int \frac{129}{x - 1} d x - \int \frac{72}{( x - 1 ) ^{2}} d x - \int \frac{30}{( x - 1 ) ^{3}} d x - \int \frac{8}{( x - 1 ) ^{4}} d x - \int \frac{1}{( x - 1 ) ^{5}} d x

\int \frac{128}{x - 2} d x = 128 ln ∣ x - 2 ∣

\int \frac{64}{( x - 2 ) ^{2}} d x = - \frac{64}{x - 2}

\int \frac{129}{x - 1} d x = 129 ln ∣ x - 1 ∣

\int \frac{72}{( x - 1 ) ^{2}} d x = - \frac{72}{x - 1}

\int \frac{30}{( x - 1 ) ^{3}} d x = - \frac{15}{( x - 1 ) ^{2}}

\int \frac{8}{( x - 1 ) ^{4}} d x = - \frac{8}{3 ( x - 1 ) ^{3}}

\int \frac{1}{( x - 1 ) ^{5}} d x = - \frac{1}{4 ( x - 1 ) ^{4}}

= 128 ln ∣ x - 2 ∣ - (- \frac{64}{x - 2}) - 129 ln ∣ x - 1 ∣ - (- \frac{72}{x - 1}) - (- \frac{15}{( x - 1 ) ^{2}}) - (- \frac{8}{3 ( x - 1 ) ^{3}}) - (- \frac{1}{4 ( x - 1 ) ^{4}})

簡素化

= 128 ln ∣ x - 2 ∣ + \frac{64}{x - 2} - 129 ln ∣ x - 1 ∣ + \frac{72}{x - 1} + \frac{15}{( x - 1 ) ^{2}} + \frac{8}{3 ( x - 1 ) ^{3}} + \frac{1}{4 ( x - 1 ) ^{4}}

定数を解答に追加する

= 128 ln ∣ x - 2 ∣ + \frac{64}{x - 2} - 129 ln ∣ x - 1 ∣ + \frac{72}{x - 1} + \frac{15}{( x - 1 ) ^{2}} + \frac{8}{3 ( x - 1 ) ^{3}} + \frac{1}{4 ( x - 1 ) ^{4}} + C