integral of (\sqrt[3]{x}+1)/(sqrt(x)+1)

解

\int \frac{3 x + 1}{x + 1} d x

\frac{6}{5} x^{\frac{5}{6}} - 3 3 x - 2 ln 6 x + 1 + ln 4 3 x - 4 6 x + 4 + 23 arctan (\frac{1}{3} (2 6 x - 1)) + 2 (x + 1 - ln x + 1) + C

解答ステップ

\int \frac{3 x + 1}{x + 1} d x

拡張

\frac{3 x + 1}{x + 1} : \frac{3 x}{x + 1} + \frac{1}{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{3 x}{x + 1} d x + \int \frac{1}{x + 1} d x

\int \frac{3 x}{x + 1} d x = \frac{6}{5} x^{\frac{5}{6}} - 3 3 x - 2 ln 6 x + 1 + ln 4 3 x - 4 6 x + 4 + 23 arctan (\frac{1}{3} (2 6 x - 1))

\int \frac{1}{x + 1} d x = 2 (x + 1 - ln x + 1)

= \frac{6}{5} x^{\frac{5}{6}} - 3 3 x - 2 ln 6 x + 1 + ln 4 3 x - 4 6 x + 4 + 23 arctan (\frac{1}{3} (2 6 x - 1)) + 2 (x + 1 - ln x + 1)

定数を解答に追加する

= \frac{6}{5} x^{\frac{5}{6}} - 3 3 x - 2 ln 6 x + 1 + ln 4 3 x - 4 6 x + 4 + 23 arctan (\frac{1}{3} (2 6 x - 1)) + 2 (x + 1 - ln x + 1) + C