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

解

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

4 x^{2} + 2 x - 3 ln x^{2} + 2 x + x + 1 + C

解答ステップ

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

拡張

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

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

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

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

= 4 (x^{2} + 2 x - ln x^{2} + 2 x + x + 1) + ln x^{2} + 2 x + x + 1

簡素化

4 (x^{2} + 2 x - ln x^{2} + 2 x + x + 1) + ln x^{2} + 2 x + x + 1 : 4 x^{2} + 2 x - 3 ln x^{2} + 2 x + x + 1

= 4 x^{2} + 2 x - 3 ln x^{2} + 2 x + x + 1

定数を解答に追加する

= 4 x^{2} + 2 x - 3 ln x^{2} + 2 x + x + 1 + C