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

解

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

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

解答ステップ

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

拡張

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

\int \frac{x}{3 x + 1} d x = \frac{3}{5} x^{\frac{5}{3}} - \frac{3}{4} x^{\frac{4}{3}} + x - \frac{3}{2} x^{\frac{2}{3}} + 3 3 x - 3 ln 3 x + 1

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

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

簡素化

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

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

定数を解答に追加する

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