integral of (x^2+4x+7)/(x^2-6x+5)

解

\int \frac{x ^{2} + 4 x + 7}{x ^{2} - 6 x + 5} d x

\frac{1}{2} (6 ln \frac{1}{4} (x - 3)^{2} - 1 - 13 (\frac{ln \frac{1}{2} ( x - 3 ) + 1}{2} - \frac{ln \frac{1}{2} ( x - 3 ) - 1}{2})) + x + 2 ln x^{2} - 6 x + 5 - 3 ln \frac{1}{2} (x - 3) + 1 + 3 ln \frac{1}{2} (x - 3) - 1 + 7 (- \frac{1}{4} ln ∣ x - 1 ∣ + \frac{1}{4} ln ∣ x - 5 ∣) + C

解答ステップ

\int \frac{x ^{2} + 4 x + 7}{x ^{2} - 6 x + 5} d x

拡張

\frac{x ^{2} + 4 x + 7}{x ^{2} - 6 x + 5} : \frac{x ^{2}}{x ^{2} - 6 x + 5} + \frac{4 x}{x ^{2} - 6 x + 5} + \frac{7}{x ^{2} - 6 x + 5}

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{x ^{2}}{x ^{2} - 6 x + 5} d x + \int \frac{4 x}{x ^{2} - 6 x + 5} d x + \int \frac{7}{x ^{2} - 6 x + 5} d x

\int \frac{x ^{2}}{x ^{2} - 6 x + 5} d x = \frac{1}{2} (6 ln \frac{1}{4} (x - 3)^{2} - 1 - 13 (\frac{ln \frac{1}{2} ( x - 3 ) + 1}{2} - \frac{ln \frac{1}{2} ( x - 3 ) - 1}{2})) + x

\int \frac{4 x}{x ^{2} - 6 x + 5} d x = 2 ln x^{2} - 6 x + 5 - 3 ln \frac{1}{2} (x - 3) + 1 + 3 ln \frac{1}{2} (x - 3) - 1

\int \frac{7}{x ^{2} - 6 x + 5} d x = 7 (- \frac{1}{4} ln ∣ x - 1 ∣ + \frac{1}{4} ln ∣ x - 5 ∣)

= \frac{1}{2} (6 ln \frac{1}{4} (x - 3)^{2} - 1 - 13 (\frac{ln \frac{1}{2} ( x - 3 ) + 1}{2} - \frac{ln \frac{1}{2} ( x - 3 ) - 1}{2})) + x + 2 ln x^{2} - 6 x + 5 - 3 ln \frac{1}{2} (x - 3) + 1 + 3 ln \frac{1}{2} (x - 3) - 1 + 7 (- \frac{1}{4} ln ∣ x - 1 ∣ + \frac{1}{4} ln ∣ x - 5 ∣)

定数を解答に追加する

= \frac{1}{2} (6 ln \frac{1}{4} (x - 3)^{2} - 1 - 13 (\frac{ln \frac{1}{2} ( x - 3 ) + 1}{2} - \frac{ln \frac{1}{2} ( x - 3 ) - 1}{2})) + x + 2 ln x^{2} - 6 x + 5 - 3 ln \frac{1}{2} (x - 3) + 1 + 3 ln \frac{1}{2} (x - 3) - 1 + 7 (- \frac{1}{4} ln ∣ x - 1 ∣ + \frac{1}{4} ln ∣ x - 5 ∣) + C