integral of (5x^2+15x+8)/((x+2)^2x)

解

\int \frac{5 x ^{2} + 15 x + 8}{( x + 2 ) ^{2} x} d x

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

解答ステップ

\int \frac{5 x ^{2} + 15 x + 8}{( x + 2 ) ^{2} x} d x

拡張

\frac{5 x ^{2} + 15 x + 8}{( x + 2 ) ^{2} x} : \frac{5 x ^{2}}{( x + 2 ) ^{2} x} + \frac{15 x}{( x + 2 ) ^{2} x} + \frac{8}{( x + 2 ) ^{2} 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{5 x ^{2}}{( x + 2 ) ^{2} x} d x + \int \frac{15 x}{( x + 2 ) ^{2} x} d x + \int \frac{8}{( x + 2 ) ^{2} x} d x

\int \frac{5 x ^{2}}{( x + 2 ) ^{2} x} d x = 5 (ln ∣ x + 2 ∣ + \frac{2}{x + 2})

\int \frac{15 x}{( x + 2 ) ^{2} x} d x = - \frac{15}{x + 2}

\int \frac{8}{( x + 2 ) ^{2} x} d x = 8 (- \frac{1}{4} ln ∣ x + 2 ∣ + \frac{1}{2 ( x + 2 )} + \frac{1}{4} ln ∣ x ∣)

= 5 (ln ∣ x + 2 ∣ + \frac{2}{x + 2}) - \frac{15}{x + 2} + 8 (- \frac{1}{4} ln ∣ x + 2 ∣ + \frac{1}{2 ( x + 2 )} + \frac{1}{4} ln ∣ x ∣)

簡素化

5 (ln ∣ x + 2 ∣ + \frac{2}{x + 2}) - \frac{15}{x + 2} + 8 (- \frac{1}{4} ln ∣ x + 2 ∣ + \frac{1}{2 ( x + 2 )} + \frac{1}{4} ln ∣ x ∣) : 3 ln ∣ x + 2 ∣ - \frac{1}{x + 2} + 2 ln ∣ x ∣

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

定数を解答に追加する

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