Question

como fazer essa integral?

$\int\limits {\frac{2}{2x^2+3x+1}} \, dx$

sei que e integral por fraçoes parciais mas ta tenso

Answer 1

∫ 2/(2x²+3x+1) dx

ax²+bx+c=a*(x-x')*(x-x'')

2x²+3x+1=0

x'=-1

x''=-1/2

2x²+3x+1 =2*(x+1)*(x+1/2)

∫ 2/2*(x+1)*(x+1/2) dx

∫ 1/(x+1)*(x+1/2) dx

1/(x+1)*(x+1/2) = A/(x+1) + B/(x+1/2)

1/(x+1)*(x+1/2) = [A(x+1/2) + B(x+1)]/(x+1)(x+1/2)

1 = A(x+1/2) + B(x+1)

0*x + 1=x*(A+B) + A/2+B

A+B=0 ==> A=-B

1= A/2 + B ==> 1= -B/2 + B ==> B =2 e A = -2

∫ -2/(x+1) + 2/(x+1/2) dx

∫ -2/(x+1) dx + ∫2/(x+1/2) dx

∫ -2/(x+1) dx ...fazendo u =x+1 ==> du =dx

∫ -2/u du = - 2 * ln(x+1) + c'

∫2/(x+1/2) dx ..fazendo u =x+1/2 ==>du=dx

∫2/(u) du =2 * ln(x+1/2) +c''

c=c'+c''

∫ -2/(x+1) dx + ∫2/(x+1/2) dx = - 2 * ln(x+1) + 2 * ln(x+1/2) + c