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

解

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

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

解答ステップ

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

拡張

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

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

\int t d t = \frac{t ^{2}}{2}

\int 3 d t = 3 t

\int \frac{3}{t} d t = 3 ln ∣ t ∣

\int \frac{1}{t ^{2}} d t = - \frac{1}{t}

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

定数を解答に追加する

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