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

解

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

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

解答ステップ

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

拡張

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

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

\int \frac{2 x ^{2}}{( x - 3 ) ^{2}} d x = 2 (- \frac{x ^{2}}{x - 3} + 2 (x - 3 + 3 ln ∣ x - 3 ∣))

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

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

定数を解答に追加する

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