integral of (2x-1)/(sqrt(x^2+9))

解

\int \frac{2 x - 1}{x ^{2} + 9} d x

2 x^{2} + 9 - ln (\frac{1}{3} x + 9 + x^{2}) + C

解答ステップ

\int \frac{2 x - 1}{x ^{2} + 9} d x

拡張

\frac{2 x - 1}{x ^{2} + 9} : \frac{2 x}{x ^{2} + 9} - \frac{1}{x ^{2} + 9}

= \int \frac{2 x}{x ^{2} + 9} - \frac{1}{x ^{2} + 9} 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}{x ^{2} + 9} d x - \int \frac{1}{x ^{2} + 9} d x

\int \frac{2 x}{x ^{2} + 9} d x = 2 x^{2} + 9

\int \frac{1}{x ^{2} + 9} d x = ln (\frac{1}{3} x + 9 + x^{2})

= 2 x^{2} + 9 - ln (\frac{1}{3} x + 9 + x^{2})

定数を解答に追加する

= 2 x^{2} + 9 - ln (\frac{1}{3} x + 9 + x^{2}) + C