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

解

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

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

解答ステップ

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

拡張

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

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

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

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

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

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

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

簡素化

- \frac{t ^{2}}{2 ( t + 3 ) ^{2}} + ln ∣ t + 3 ∣ + \frac{3}{t + 3} - \frac{1}{t + 3} + \frac{3}{2 ( t + 3 ) ^{2}} - \frac{1}{2 ( t + 3 ) ^{2}} : - \frac{t ^{2}}{2 ( t + 3 ) ^{2}} + ln ∣ t + 3 ∣ + \frac{2}{t + 3} + \frac{1}{( t + 3 ) ^{2}}

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

定数を解答に追加する

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