integral of (2x+7)/(sqrt(x^2+5x-4))

解

\int \frac{2 x + 7}{x ^{2} + 5 x - 4} d x

2 x^{2} + 5 x - 4 + 2 ln \frac{1}{41} (4 x^{2} + 20 x - 16) + \frac{1}{41} (2 x + 5) + C

解答ステップ

\int \frac{2 x + 7}{x ^{2} + 5 x - 4} d x

拡張

\frac{2 x + 7}{x ^{2} + 5 x - 4} : \frac{2 x}{x ^{2} + 5 x - 4} + \frac{7}{x ^{2} + 5 x - 4}

= \int \frac{2 x}{x ^{2} + 5 x - 4} + \frac{7}{x ^{2} + 5 x - 4} 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} + 5 x - 4} d x + \int \frac{7}{x ^{2} + 5 x - 4} d x

\int \frac{2 x}{x ^{2} + 5 x - 4} d x = 2 x^{2} + 5 x - 4 - 5 ln \frac{1}{41} (4 x^{2} + 20 x - 16) + \frac{2}{41} x + \frac{5}{41}

\int \frac{7}{x ^{2} + 5 x - 4} d x = 7 ln \frac{1}{41} (4 x^{2} + 20 x - 16) + \frac{2}{41} x + \frac{5}{41}

= 2 x^{2} + 5 x - 4 - 5 ln \frac{1}{41} (4 x^{2} + 20 x - 16) + \frac{2}{41} x + \frac{5}{41} + 7 ln \frac{1}{41} (4 x^{2} + 20 x - 16) + \frac{2}{41} x + \frac{5}{41}

簡素化

2 x^{2} + 5 x - 4 - 5 ln \frac{1}{41} (4 x^{2} + 20 x - 16) + \frac{2}{41} x + \frac{5}{41} + 7 ln \frac{1}{41} (4 x^{2} + 20 x - 16) + \frac{2}{41} x + \frac{5}{41} : 2 x^{2} + 5 x - 4 + 2 ln \frac{1}{41} (4 x^{2} + 20 x - 16) + \frac{1}{41} (2 x + 5)

= 2 x^{2} + 5 x - 4 + 2 ln \frac{1}{41} (4 x^{2} + 20 x - 16) + \frac{1}{41} (2 x + 5)

定数を解答に追加する

= 2 x^{2} + 5 x - 4 + 2 ln \frac{1}{41} (4 x^{2} + 20 x - 16) + \frac{1}{41} (2 x + 5) + C