integral of (t+1)/(3t^2+6t-5)

解

\int \frac{t + 1}{3 t ^{2} + 6 t - 5} d t

\frac{1}{6} ln 3 t^{2} + 6 t - 5 + C

解答ステップ

\int \frac{t + 1}{3 t ^{2} + 6 t - 5} d t

拡張

\frac{t + 1}{3 t ^{2} + 6 t - 5} : \frac{t}{3 t ^{2} + 6 t - 5} + \frac{1}{3 t ^{2} + 6 t - 5}

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}{3 t ^{2} + 6 t - 5} d t + \int \frac{1}{3 t ^{2} + 6 t - 5} d t

\int \frac{t}{3 t ^{2} + 6 t - 5} d t = \frac{1}{6} ln 3 t^{2} + 6 t - 5 + \frac{1}{4 6} (ln \frac{3}{2 2} (t + 1) + 1 - ln \frac{3}{2 2} (t + 1) - 1)

\int \frac{1}{3 t ^{2} + 6 t - 5} d t = - \frac{1}{4 6} (ln \frac{3}{2 2} (t + 1) + 1 - ln \frac{3}{2 2} (t + 1) - 1)

= \frac{1}{6} ln 3 t^{2} + 6 t - 5 + \frac{1}{4 6} (ln \frac{3}{2 2} (t + 1) + 1 - ln \frac{3}{2 2} (t + 1) - 1) - \frac{1}{4 6} (ln \frac{3}{2 2} (t + 1) + 1 - ln \frac{3}{2 2} (t + 1) - 1)

類似した元を足す:

\frac{1}{4 6} (ln \frac{3}{2 2} (t + 1) + 1 - ln \frac{3}{2 2} (t + 1) - 1) - \frac{1}{4 6} (ln \frac{3}{2 2} (t + 1) + 1 - ln \frac{3}{2 2} (t + 1) - 1) = 0

= \frac{1}{6} ln 3 t^{2} + 6 t - 5

定数を解答に追加する

= \frac{1}{6} ln 3 t^{2} + 6 t - 5 + C