intergral 2x^2/(x^2+4x+8)
I did a long division and got
2x^2/(x^2+4x+8) = 2 - (4x+8)/(x^2 + 4x + 8)
= 2 - 2(2x+4)/(x^2 + 4x + 8)
I noticed that the derivative of x^2 + 4x + 8 is 2x+4 , which I have sitting on top.
Ahhh, logs!!!!
∫2x^2/(x^2+4x+8) dx
=∫ ( 2 - 2(2x+4)/x^2 + 4x + 8) ) dx
= 2x - 4 ln(x^2 + 4x + 8)
To solve the integral of 2x^2 / (x^2 + 4x + 8), we can use a technique called partial fraction decomposition.
Step 1: Factorize the denominator:
x^2 + 4x + 8 = (x + 2)(x + 2)
Step 2: Express the fraction in terms of partial fractions:
2x^2 / (x^2 + 4x + 8) = A / (x + 2) + B / (x + 2)
Step 3: Multiply through by (x^2 + 4x + 8) to remove the denominators:
2x^2 = A(x + 2) + B(x + 2)
Step 4: Expand and equate like terms:
2x^2 = (A + B)x + 2(A + B)
Step 5: Solve for A and B by equating coefficients:
2x^2 coefficients, we get: A + B = 2
Constant coefficients, we get: 2(A + B) = 0
Step 6: Solve the system of equations to find A and B:
From the second equation, we find: A + B = 0 (dividing by 2)
Substituting this value in the first equation, we have: 0 = 2, which doesn't hold true.
Since we cannot find specific values for A and B, we can conclude that the original integral cannot be solved using partial fraction decomposition.