integral of (8+t+6t^2-12t^3)/((3t^2+4)(t^2+7))

解

\int \frac{8 + t + 6 t ^{2} - 12 t ^{3}}{( 3 t ^{2} + 4 ) ( t ^{2} + 7 )} d t

\frac{1}{2} ln 3 t^{2} + 4 - \frac{5}{2} ln t^{2} + 7 + \frac{2}{7} arctan (\frac{t}{7}) + C

解答ステップ

\int \frac{8 + t + 6 t ^{2} - 12 t ^{3}}{( 3 t ^{2} + 4 ) ( t ^{2} + 7 )} d t

以下の部分分数を得る:

\frac{8 + t + 6 t ^{2} - 12 t ^{3}}{( 3 t ^{2} + 4 ) ( t ^{2} + 7 )} : \frac{3 t}{3 t ^{2} + 4} + \frac{- 5 t + 2}{t ^{2} + 7}

= \int \frac{3 t}{3 t ^{2} + 4} + \frac{- 5 t + 2}{t ^{2} + 7} 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{3 t}{3 t ^{2} + 4} d t + \int \frac{- 5 t + 2}{t ^{2} + 7} d t

\int \frac{3 t}{3 t ^{2} + 4} d t = \frac{1}{2} ln 3 t^{2} + 4

\int \frac{- 5 t + 2}{t ^{2} + 7} d t = - \frac{5}{2} ln t^{2} + 7 + \frac{2}{7} arctan (\frac{t}{7})

= \frac{1}{2} ln 3 t^{2} + 4 - \frac{5}{2} ln t^{2} + 7 + \frac{2}{7} arctan (\frac{t}{7})

定数を解答に追加する

= \frac{1}{2} ln 3 t^{2} + 4 - \frac{5}{2} ln t^{2} + 7 + \frac{2}{7} arctan (\frac{t}{7}) + C