Question

Вычислить неопределенные интегралы: A) { sinx dx/ 1+2cosx B) {dx/x(x+1) C) {(x-1)e^x dx Попроси больше объяснений Следить Отметить нарушение Лесенок5 13.04.2016 Войти чтобы добавить комментарий ОтветПроверено экспертомОтвет данNNNLLL54

Answer 1

A) ∫(sinx dx / 1 + 2cosx) To solve this integral, we can use the substitution method. Let's substitute u = tan(x/2), then we have du = (1/2)(sec^2(x/2)) dx. We also know that sinx = 2tan(x/2) / (1 + tan^2(x/2)), and cosx = (1 - tan^2(x/2)) / (1 + tan^2(x/2)).

Using these substitutions, we can rewrite the integral as:

∫(2tan(x/2) / (1 + tan^2(x/2))) * (1/2)(sec^2(x/2)) dx

= ∫(du / u^2 + 1)

This integral is a standard form and equals arctan(u) + C. Substituting back u = tan(x/2), we get:

arctan(tan(x/2)) + C

B) ∫(dx / x(x+1)) To solve this integral, we can use partial fraction decomposition. We can rewrite the integrand as:

1/x(x+1) = A/x + B/(x+1)

Multiplying through by x(x+1), we get:

1 = A(x+1) + Bx

Setting x = 0, we get A = 1. Setting x = -1, we get B = -1.

So the integral becomes:

∫(dx / x(x+1)) = ∫(1/x - 1/(x+1)) dx

= ln|x| - ln|x+1| + C

C) ∫((x-1)e^x dx) To solve this integral, we can use integration by parts. Let u = x-1 and dv = e^x dx. Then du = dx and v = e^x.

Using the formula for integration by parts:

∫u dv = uv - ∫v du

We get:

(x-1)e^x - ∫e^x dx

= (x-1)e^x - e^x + C

So the indefinite integral is:

(x-1)e^x - e^x + C