Saved Bookmarks
| 1. |
Find integral of x upon x3 + x2 + x + 1.\(\int \frac {x\,dx}{x^3+x^2+x+1} \) |
|
Answer» \(\int \frac {x\,dx}{x^3+x^2+x+1} = \int \frac {x\, dx}{(x^2+1) (x+1)}\) Let \(\frac {x}{(x^2+1) (x+1)} = \frac {A}{x+1} + \frac {Bx+C}{x^2+1}\) ∴ x = A (x2+1) + (Bx + C) (x + 1) Put x = -1, we get 2A = -1 = A = -1/2 Put x = 0, We get A + C = 0 = C = A = 1/2 Put x = 1, We get 2A + 2B + 2C = 1 = 2B = 1-2A-2C = 1 + 1 -1 = 1 = B = 1/2 ∴ \(\frac {x}{(x^2+1)(x+1)} = \frac {-1}{2} \frac {1}{x+1} + \frac 12 \frac {1}{x^2+1} + \frac 12 \frac {1}{x^2+1}\) = \(\int \frac {x}{(x^2+1)(x+1)}dx = \frac {-1}{2} \int \frac {1}{x+1}dx + \frac 14 \int \frac {2x}{x^2+1}dx+\frac 12 \int \frac {1}{x^2+1}dx\) = \(\frac {-1}{2} log [x+1] + \frac 14lg [x^2+1] + \frac 12 ta\vec n \,x+C\) = \(\frac {1}{4} log (\frac {x^2+1}{(x+1)^2}) + \frac 12 ta\vec n \,x+C\) |
|