integral of ((4x-1))/(\sqrt[3]{3x+1)}

解

\int \frac{( 4 x - 1 )}{3 3 x + 1} d x

4 (\frac{1}{2} x (3 x + 1)^{\frac{2}{3}} - \frac{1}{10} (3 x + 1)^{\frac{5}{3}}) - \frac{1}{2} (3 x + 1)^{\frac{2}{3}} + C

解答ステップ

\int \frac{4 x - 1}{3 3 x + 1} d x

拡張

\frac{4 x - 1}{3 3 x + 1} : \frac{4 x}{3 3 x + 1} - \frac{1}{3 3 x + 1}

= \int \frac{4 x}{3 3 x + 1} - \frac{1}{3 3 x + 1} d x

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

\int \frac{4 x}{3 3 x + 1} d x = 4 (\frac{1}{2} x (3 x + 1)^{\frac{2}{3}} - \frac{1}{10} (3 x + 1)^{\frac{5}{3}})

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

= 4 (\frac{1}{2} x (3 x + 1)^{\frac{2}{3}} - \frac{1}{10} (3 x + 1)^{\frac{5}{3}}) - \frac{1}{2} (3 x + 1)^{\frac{2}{3}}

定数を解答に追加する

= 4 (\frac{1}{2} x (3 x + 1)^{\frac{2}{3}} - \frac{1}{10} (3 x + 1)^{\frac{5}{3}}) - \frac{1}{2} (3 x + 1)^{\frac{2}{3}} + C