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

解

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

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

解答ステップ

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

拡張

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

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

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

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

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

定数を解答に追加する

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