integral of t(1-ln^2(t))

解

\int t (1 - ln^{2} (t)) d t

\frac{t ^{2}}{2} - \frac{1}{2} t^{2} ln^{2} (t) + \frac{1}{2} t^{2} ln (t) - \frac{t ^{2}}{4} + C

解答ステップ

\int t (1 - ln^{2} (t)) d t

拡張

t (1 - ln^{2} (t)) : t - t ln^{2} (t)

= \int t - t ln^{2} (t) 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 t ln^{2} (t) d t

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

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

= \frac{t ^{2}}{2} - (\frac{1}{2} t^{2} ln^{2} (t) - \frac{1}{2} t^{2} ln (t) + \frac{t ^{2}}{4})

簡素化

= \frac{t ^{2}}{2} - \frac{1}{2} t^{2} ln^{2} (t) + \frac{1}{2} t^{2} ln (t) - \frac{t ^{2}}{4}

定数を解答に追加する

= \frac{t ^{2}}{2} - \frac{1}{2} t^{2} ln^{2} (t) + \frac{1}{2} t^{2} ln (t) - \frac{t ^{2}}{4} + C