integral of (2x+1)/(sqrt(4x-1))

解

\int \frac{2 x + 1}{4 x - 1} d x

\frac{1}{3} x 4 x - 1 + \frac{2}{3} 4 x - 1 + C

解答ステップ

\int \frac{2 x + 1}{4 x - 1} d x

拡張

\frac{2 x + 1}{4 x - 1} : \frac{2 x}{4 x - 1} + \frac{1}{4 x - 1}

= \int \frac{2 x}{4 x - 1} + \frac{1}{4 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{2 x}{4 x - 1} d x + \int \frac{1}{4 x - 1} d x

\int \frac{2 x}{4 x - 1} d x = \frac{1}{6} (2 x + 1) 4 x - 1

\int \frac{1}{4 x - 1} d x = \frac{1}{2} 4 x - 1

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

簡素化

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

= \frac{1}{3} x 4 x - 1 + \frac{2}{3} 4 x - 1

定数を解答に追加する

= \frac{1}{3} x 4 x - 1 + \frac{2}{3} 4 x - 1 + C