integral of (4/(x^3)+2)/(1/(x^2)-x)

解

\int \frac{\frac{4}{x ^{3}} + 2}{\frac{1}{x ^{2}} - x} d x

\frac{4}{3} (ln x^{3} - ln x^{3} - 1) - \frac{2}{3} ln 1 - x^{3} + C

解答ステップ

\int \frac{\frac{4}{x ^{3}} + 2}{\frac{1}{x ^{2}} - x} d x

拡張

\frac{\frac{4}{x ^{3}} + 2}{\frac{1}{x ^{2}} - x} : \frac{\frac{4}{x ^{3}}}{\frac{1}{x ^{2}} - x} + \frac{2}{\frac{1}{x ^{2}} - 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{\frac{4}{x ^{3}}}{\frac{1}{x ^{2}} - x} d x + \int \frac{2}{\frac{1}{x ^{2}} - x} d x

\int \frac{\frac{4}{x ^{3}}}{\frac{1}{x ^{2}} - x} d x = \frac{4}{3} (ln x^{3} - ln x^{3} - 1)

\int \frac{2}{\frac{1}{x ^{2}} - x} d x = - \frac{2}{3} ln 1 - x^{3}

= \frac{4}{3} (ln x^{3} - ln x^{3} - 1) - \frac{2}{3} ln 1 - x^{3}

定数を解答に追加する

= \frac{4}{3} (ln x^{3} - ln x^{3} - 1) - \frac{2}{3} ln 1 - x^{3} + C