dx 1 x4 Solution 1x4 1 dx very tough integral the denomin

dx / 1 + x4

Solution

? [1/(x4+ 1)] dx very tough integral...! the denominator is a sum of squares, thus it is not factorable, therefore partial fraction method seems inapplicable; nevertheless, you can apply it on condition that you rearrange the integrand as follows: add and subtract 2x² at denominator, in order to complete the square: ? [1/ (x4+ 2x² +1 - 2x²)] dx = ? {1/ [(x4+ 2x² +1) - 2x²]} dx = ? {1/ [(x² +1)² - 2x²]} dx = now factor the denominator, viewing it as a difference between two squares: ? {1/ [(x² +1)² - (v2x)²]} dx = ? {1/ {[(x² +1) + (v2x)] [(x² +1) - (v2x)]}} dx = ? {1/ [(x² + v2x +1)(x² - v2x +1)]} dx = well, this rearranged integrand is solvable by partial fractions: so you have to find four real numbers A, B, C and D, so that: 1/ [(x² +v2x +1)(x² -v2x +1)] = (Ax + B) /(x² +v2x +1) + (Cx + D) /(x² -v2x +1) ? 1/ [(x² +v2x +1)(x² -v2x +1)] = [(Ax + B)(x² -v2x +1) + (Cx + D)(x² +v2x +1)] / [(x² +v2x +1)(x² -v2x +1)] ? equating the numerators: 1 = [(Ax + B)(x² -v2x +1) + (Cx + D)(x² +v2x +1)] ? 1 = Ax³ -v2Ax² + Ax + Bx² -v2Bx + B + Cx³+ v2Cx²+ Cx + Dx² +v2Dx + D ? 1 = (A + C)x³ + (-v2A + B + v2C + D)x²+ (A -v2B + C +v2D)x + (B + D) ? leading to the system: | A + C = 0 ? A = - C ? A = 1/(2v2) | -v2A + B +v2C+ D = 0 ? -v2(- C)+ B +v2C+ D = 0 ? 2v2C + 1 = 0 ? C = -1/(2v2) | A -v2B + C +v2D = 0 ? - C -v2B + C +v2D = 0 ? v2D = v2B ? D = B ? B = (1/2) | B + D = 1 ? D + D = 1 ? D = (1/2) now, plugging in the found values, you get (see above): 1/ [(x² +v2x +1)(x² -v2x +1)] = (Ax + B) /(x² +v2x +1) + (Cx + D) /(x² -v2x +1) ? 1/ [(x² +v2x +1)(x² -v2x +1)] = {[1/(2v2)]x + (1/2)} /(x² +v2x +1) + {[-1/(2v2)]x + (1/2)} /(x² -v2x +1) ? factoring out [1/(2v2)] at both fractions: 1/ [(x² +v2x +1)(x² -v2x +1)] = [1/(2v2)] [(x +v2) /(x² +v2x +1)] + [1/(2v2)] [(-x +v2) /(x² -v2x +1)] ? therefore: ? {1/ [(x² + v2x +1)(x² - v2x +1)]} dx = ? {[1/(2v2)] [(x +v2) /(x² +v2x +1)] + [1/(2v2)] [(-x +v2) /(x² -v2x +1)]} dx = breaking it up and taking the constants out: [1/(2v2)] ? [(x +v2) /(x² +v2x +1)] dx + [1/(2v2)] ? [(-x +v2)} /(x² -v2x +1)] dx = now, in order to change each numerator into the derivative of the respective denominator, divide and multiply the first integral by 2 and the second one by (-2), yielding: [1/(2v2)](1/2) ? [2(x+v2) /(x²+v2x+1)] dx+ [1/(2v2)](-1/2) ? [(-2)(-x+v2)} /(x²-v2x+1)] dx = [1/(4v2)] ? [(2x + 2v2) /(x²+v2x +1)] dx - [1/(4v2)] ? [(2x - 2v2)} /(x²-v2x +1)] dx = then rewrite the numerators as: [1/(4v2)] ? [(2x +v2 +v2) /(x²+v2x +1)] dx - [1/(4v2)] ? [(2x - v2 - v2)} /(x²-v2x +1)] dx = and distribute them as: [1/(4v2)] ? {[(2x +v2)/(x²+v2x +1)] + [v2 /(x²+v2x +1)]} dx - [1/(4v2)] ? {[(2x - v2)/(x²-v2x +1)] - [v2 /(x²-v2x +1)]} dx = breaking it up, [1/(4v2)] ? [(2x +v2)/(x²+v2x +1)] dx + [1/(4v2)] ? [v2 /(x²+v2x +1)] dx - [1/(4v2)] ? [(2x - v2)/(x²-v2x +1)] dx + [1/(4v2)] ? [v2 /(x²-v2x +1)] dx = simplifying; [1/(4v2)] ? [d(x²+v2x +1)] /(x²+v2x +1) + (1/4) ? [1 /(x²+v2x +1)] dx - [1/(4v2)] ? [d(x²-v2x +1)] /(x²-v2x +1)] + (1/4) ? [1 /(x²-v2x +1)] dx = [1/(4v2)] ln (x²+v2x +1) + (1/4) ? [1 /(x²+v2x +1)] dx - [1/(4v2)] ln (x²-v2x +1) + (1/4) ? [1 /(x²-v2x +1)] dx = [1/(4v2)] [ln (x²+v2x +1) - ln (x²-v2x +1)] + (1/4) ? [1 /(x²+v2x +1)] dx + (1/4) ? [1 /(x²-v2x +1)] dx = according to log properties, [1/(4v2)] ln [(x²+v2x +1)/(x²-v2x +1)] + (1/4) ? [1/(x²+v2x +1)] dx + (1/4) ? [1/(x²-v2x +1)] dx = to solve the remaining integrals, you need to complete the squares at denominators, since they aren\'t factorable; more exactly, you have to attempt to change the denominators into {[f (x)]²+ 1} form (which is then integrable as arctan: ? {d[f(x)]} /{[f(x)]² + 1} = arctan[f(x)] + c); thus, first, multiply and divide the remaining integrals by 2: [1/(4v2)] ln [(x²+v2x +1)/(x²-v2x +1)] + (1/4)(2) ? dx/[2(x²+v2x +1)] + (1/4)(2) ? dx/ [2(x²-v2x +1)] = [1/(4v2)] ln [(x²+v2x +1)/(x²-v2x +1)] + (1/2) ? dx/(2x²+2v2x +2) + (1/2) ? dx/ (2x²-2v2x +2) = (rewriting 2 as 1+1) [1/(4v2)] ln [(x²+v2x +1)/(x²-v2x +1)] + (1/2) ? dx/(2x² + 2v2x +1 +1) + (1/2) ? dx/ (2x²- 2v2x +1 + 1) = [1/(4v2)] ln [(x²+v2x +1)/(x²-v2x +1)] + (1/2) ? dx/[(2x² + 2v2x +1) +1] + (1/2) ? dx/ [(2x²- 2v2x +1) +1] = having completed the squares, [1/(4v2)] ln [(x²+v2x +1)/(x²-v2x +1)] + (1/2) ? dx/[(v2x +1)²+1] + (1/2) ? dx/ [(v2x -1)²+1] = finally, divide and multiply the integrals by v2: [1/(4v2)] ln [(x²+v2x +1)/(x²-v2x +1)] + (1/2)(1/v2) ? (v2dx)/[(v2x +1)² +1] + (1/2)(1/v2) ? (v2dx)/ [(v2x -1)² +1] = so that you can rewrite the numerators as: [1/(4v2)] ln [(x²+v2x +1)/(x²-v2x +1)] + [1/(2v2)] ? [d(v2x +1)] /[(v2x +1)² +1] + [1/(2v2)] ? [d(v2x - 1)] / [(v2x -1)² +1] = [1/(4v2)] ln [(x²+v2x +1)/(x²-v2x +1)] + [1/(2v2)] arctan (v2x +1) + [1/(2v2)] arctan (v2x - 1) + c in conclusion, ? [1/(x4+ 1)] dx = [1/(4v2)] ln [(x²+v2x +1)/(x²-v2x +1)] + [1/(2v2)] arctan (v2x +1) + [1/(2v2)] arctan (v2x - 1) + c the next one is easier, and since you were able to solve it yourself, I won\'t show its solution, which needs factoring (x4-1) completely and then proceeding with usual partial fraction decomposition I hope it has been helpful... Bye and good luck!