integral of (5x-4)/(sqrt(8x-5^2+9))

解

\int \frac{5 x - 4}{8 x - 5 ^{2} + 9} d x

\frac{5}{12} x 8 x - 16 + \frac{2}{3} 8 x - 16 + C

解答ステップ

\int \frac{5 x - 4}{8 x - 5 ^{2} + 9} d x

拡張

\frac{5 x - 4}{8 x - 5 ^{2} + 9} : \frac{5 x}{8 x - 16} - \frac{4}{8 x - 16}

= \int \frac{5 x}{8 x - 16} - \frac{4}{8 x - 16} 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{5 x}{8 x - 16} d x - \int \frac{4}{8 x - 16} d x

\int \frac{5 x}{8 x - 16} d x = \frac{5}{12} (x + 4) 8 x - 16

\int \frac{4}{8 x - 16} d x = 8 x - 16

= \frac{5}{12} (x + 4) 8 x - 16 - 8 x - 16

簡素化

\frac{5}{12} (x + 4) 8 x - 16 - 8 x - 16 : \frac{5}{12} x 8 x - 16 + \frac{2}{3} 8 x - 16

= \frac{5}{12} x 8 x - 16 + \frac{2}{3} 8 x - 16

定数を解答に追加する

= \frac{5}{12} x 8 x - 16 + \frac{2}{3} 8 x - 16 + C