Вычислить неопределенный интеграл (с объяснением)
[tex] \int\limits { \frac{(2x+1)}{(x-1)^2(x^2+2x+3)} } \, dx [/tex]
[tex] \int\limits { \frac{(2x+1)}{(x-1)^2(x^2+2x+3)} } \, dx [/tex]
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Verified answer
Метод неопределенных коэффициентов(2x+1)/[(x-1)^2*(x^2+2x+3)] = A1/(x-1) + A2/(x-1)^2 + (A3*x+A4)/(x^2+2x+3) =
= [A1*(x-1)(x^2+2x+3) + A2*(x^2+2x+3) + (A3*x+A4)(x-1)^2] / [(x-1)^2*(x^2+2x+3)] =
= [A1(x^3+x^2+x-3)+A2(x^2+2x+3)+A3*x(x^2-2x+1)+A4(x^2-2x+1)] /
/ [(x-1)^2*(x^2+2x+3)] =
= [x^3(A1+A3)+x^2(A1+A2-2A3+A4)+x(A1+2A2+A3-2A4)+(-3A1+3A2+A4)] /
/ [(x-1)^2*(x^2+2x+3)] = (2x+1)/[(x-1)^2*(x^2+2x+3)]
Система
{ A1 + A3 = 0
{ A1 + A2 - 2A3 + A4 = 0
{ A1 + 2A2 + A3 - 2A4 = 2
{ -3A1 + 3A2 + A4 = 1
{ A3 = -A1
{ A1 + A2 + 2A1 + A4 = 0
{ 2A2 - 2A4 = 2
{ -3A1 + 3A2 + A4 = 1
{ A3 = -A1
{ A4 = A2 - 1
{ 3A1 + A2 + A2 - 1 = 0
{ -3A1 + 3A2 + A2 - 1 = 1
{ A3 = -A1
{ A4 = A2 - 1
{ 3A1 + 2A2 = 1
{ -3A1 + 4A2 = 2
Складываем 3 и 4 уравнения
6A2 = 3, A2 = 1/2, A4 = 1/2 - 1 = -1/2
3A1 + 2*1/2 = 1, A1 = 0, A3 = 0
Подставляем обратно в интеграл
Int (2x+1)/[(x-1)^2*(x^2+2x+3)] dx =
= Int [1/2*1/(x-1)^2 - 1/2*1/(x^2+2x+3)] dx =
= 1/2*Int 1/(x-1)^2 dx - 1/2*Int 1/((x+1)^2+2) dx =
= -1/2*1/(x-1) - 1/2*1/sqrt(2)*arctg [(x+1)/sqrt(2)] + C