integral of (x^2-3x+2)/(x^2-9)

Lời Giải

\int \frac{x ^{2} - 3 x + 2}{x ^{2} - 9} d x

x - \frac{11}{6} ln \frac{x}{3} + 1 + \frac{11}{6} ln \frac{x}{3} - 1 - \frac{3}{2} ln x^{2} - 9 + C

Các bước giải pháp

\int \frac{x ^{2} - 3 x + 2}{x ^{2} - 9} d x

Mở rộng

\frac{x ^{2} - 3 x + 2}{x ^{2} - 9} : \frac{x ^{2}}{x ^{2} - 9} - \frac{3 x}{x ^{2} - 9} + \frac{2}{x ^{2} - 9}

Áp dụng quy tắc tổng:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{x ^{2}}{x ^{2} - 9} d x - \int \frac{3 x}{x ^{2} - 9} d x + \int \frac{2}{x ^{2} - 9} d x

\int \frac{x ^{2}}{x ^{2} - 9} d x = - \frac{3}{2} ln \frac{x}{3} + 1 + \frac{3}{2} ln \frac{x}{3} - 1 + x

\int \frac{3 x}{x ^{2} - 9} d x = \frac{3}{2} ln x^{2} - 9

\int \frac{2}{x ^{2} - 9} d x = - \frac{1}{3} (ln \frac{x}{3} + 1 - ln \frac{x}{3} - 1)

= - \frac{3}{2} ln \frac{x}{3} + 1 + \frac{3}{2} ln \frac{x}{3} - 1 + x - \frac{3}{2} ln x^{2} - 9 - \frac{1}{3} (ln \frac{x}{3} + 1 - ln \frac{x}{3} - 1)

Rút gọn

- \frac{3}{2} ln \frac{x}{3} + 1 + \frac{3}{2} ln \frac{x}{3} - 1 + x - \frac{3}{2} ln x^{2} - 9 - \frac{1}{3} (ln \frac{x}{3} + 1 - ln \frac{x}{3} - 1) : x - \frac{11}{6} ln \frac{x}{3} + 1 + \frac{11}{6} ln \frac{x}{3} - 1 - \frac{3}{2} ln x^{2} - 9

= x - \frac{11}{6} ln \frac{x}{3} + 1 + \frac{11}{6} ln \frac{x}{3} - 1 - \frac{3}{2} ln x^{2} - 9

Thêm một hằng số vào Lời giải

= x - \frac{11}{6} ln \frac{x}{3} + 1 + \frac{11}{6} ln \frac{x}{3} - 1 - \frac{3}{2} ln x^{2} - 9 + C

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