integral of (7x+1)/(6x^2+x+1)

解

\int \frac{7 x + 1}{6 x ^{2} + x + 1} d x

\frac{7}{12} ln 144 x^{2} + 24 x + 24 + \frac{5}{6 23} arctan (\frac{1}{23} (12 x + 1)) + C

解答ステップ

\int \frac{7 x + 1}{6 x ^{2} + x + 1} d x

拡張

\frac{7 x + 1}{6 x ^{2} + x + 1} : \frac{7 x}{6 x ^{2} + x + 1} + \frac{1}{6 x ^{2} + 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{7 x}{6 x ^{2} + x + 1} d x + \int \frac{1}{6 x ^{2} + x + 1} d x

\int \frac{7 x}{6 x ^{2} + x + 1} d x = 14 (\frac{1}{24} ln 144 x^{2} + 24 x + 24 - \frac{1}{12 23} arctan (\frac{12 x + 1}{23}))

\int \frac{1}{6 x ^{2} + x + 1} d x = \frac{2}{23} arctan (\frac{12 x + 1}{23})

= 14 (\frac{1}{24} ln 144 x^{2} + 24 x + 24 - \frac{1}{12 23} arctan (\frac{12 x + 1}{23})) + \frac{2}{23} arctan (\frac{12 x + 1}{23})

簡素化

14 (\frac{1}{24} ln 144 x^{2} + 24 x + 24 - \frac{1}{12 23} arctan (\frac{12 x + 1}{23})) + \frac{2}{23} arctan (\frac{12 x + 1}{23}) : \frac{7}{12} ln 144 x^{2} + 24 x + 24 + \frac{5}{6 23} arctan (\frac{1}{23} (12 x + 1))

= \frac{7}{12} ln 144 x^{2} + 24 x + 24 + \frac{5}{6 23} arctan (\frac{1}{23} (12 x + 1))

定数を解答に追加する

= \frac{7}{12} ln 144 x^{2} + 24 x + 24 + \frac{5}{6 23} arctan (\frac{1}{23} (12 x + 1)) + C