integral of (2x+1)/(3x^2+8x-1)

解

\int \frac{2 x + 1}{3 x ^{2} + 8 x - 1} d x

\frac{1}{3} ln 9 x^{2} + 24 x - 3 + \frac{5}{6 19} ln \frac{1}{19} (3 x + 4) + 1 - \frac{5}{6 19} ln \frac{1}{19} (3 x + 4) - 1 + C

解答ステップ

\int \frac{2 x + 1}{3 x ^{2} + 8 x - 1} d x

拡張

\frac{2 x + 1}{3 x ^{2} + 8 x - 1} : \frac{2 x}{3 x ^{2} + 8 x - 1} + \frac{1}{3 x ^{2} + 8 x - 1}

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{2 x}{3 x ^{2} + 8 x - 1} d x + \int \frac{1}{3 x ^{2} + 8 x - 1} d x

\int \frac{2 x}{3 x ^{2} + 8 x - 1} d x = 2 (\frac{1}{6} ln 9 x^{2} + 24 x - 3 + \frac{2}{3 19} (ln \frac{3 x + 4}{19} + 1 - ln \frac{3 x + 4}{19} - 1))

\int \frac{1}{3 x ^{2} + 8 x - 1} d x = - \frac{1}{2 19} (ln \frac{3 x + 4}{19} + 1 - ln \frac{3 x + 4}{19} - 1)

= 2 (\frac{1}{6} ln 9 x^{2} + 24 x - 3 + \frac{2}{3 19} (ln \frac{3 x + 4}{19} + 1 - ln \frac{3 x + 4}{19} - 1)) - \frac{1}{2 19} (ln \frac{3 x + 4}{19} + 1 - ln \frac{3 x + 4}{19} - 1)

簡素化

2 (\frac{1}{6} ln 9 x^{2} + 24 x - 3 + \frac{2}{3 19} (ln \frac{3 x + 4}{19} + 1 - ln \frac{3 x + 4}{19} - 1)) - \frac{1}{2 19} (ln \frac{3 x + 4}{19} + 1 - ln \frac{3 x + 4}{19} - 1) : \frac{1}{3} ln 9 x^{2} + 24 x - 3 + \frac{5}{6 19} ln \frac{1}{19} (3 x + 4) + 1 - \frac{5}{6 19} ln \frac{1}{19} (3 x + 4) - 1

= \frac{1}{3} ln 9 x^{2} + 24 x - 3 + \frac{5}{6 19} ln \frac{1}{19} (3 x + 4) + 1 - \frac{5}{6 19} ln \frac{1}{19} (3 x + 4) - 1

定数を解答に追加する

= \frac{1}{3} ln 9 x^{2} + 24 x - 3 + \frac{5}{6 19} ln \frac{1}{19} (3 x + 4) + 1 - \frac{5}{6 19} ln \frac{1}{19} (3 x + 4) - 1 + C