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

解

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

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

解答ステップ

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

拡張

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

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

\int \frac{x}{x ^{2} + x - 1} d x = x^{2} + x - 1 - \frac{1}{2} ln \frac{4 x ^{2} + 4 x - 4}{5} + \frac{2}{5} x + \frac{1}{5}

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

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

簡素化

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

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

定数を解答に追加する

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