integral of (x+3)/(sqrt(x^2+6x))

解

\int \frac{x + 3}{x ^{2} + 6 x} d x

x^{2} + 6 x + C

解答ステップ

\int \frac{x + 3}{x ^{2} + 6 x} d x

拡張

\frac{x + 3}{x ^{2} + 6 x} : \frac{x}{x ^{2} + 6 x} + \frac{3}{x ^{2} + 6 x}

= \int \frac{x}{x ^{2} + 6 x} + \frac{3}{x ^{2} + 6 x} 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{x}{x ^{2} + 6 x} d x + \int \frac{3}{x ^{2} + 6 x} d x

\int \frac{x}{x ^{2} + 6 x} d x = x^{2} + 6 x - 3 ln (\frac{1}{3} x^{2} + 6 x + x + 3)

\int \frac{3}{x ^{2} + 6 x} d x = 3 ln (\frac{1}{3} x^{2} + 6 x + x + 3)

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

簡素化

= x^{2} + 6 x

定数を解答に追加する

= x^{2} + 6 x + C