Evaluate the indefinite integral of
InT(-x^3+9x^2-3x+2)/(x^4-2x^3)
I got ((1-2x)/(2x^2))+3*ln(2-x)-4*ln(x)+c
but apparently that's not the answer... and I don't know why.
See here:
http://www.jiskha.com/display.cgi?id=1316565092#1316565092.1316570036
To evaluate the indefinite integral of the given rational function, we can start by simplifying the expression and then proceed with integration.
Let's simplify the rational function first:
∫ ln(-x^3+9x^2-3x+2) / (x^4-2x^3) dx
Rewriting the denominator:
x^4 - 2x^3 = x^3(x - 2)
Now, we can express the rational function in partial fraction decomposition form:
ln(-x^3+9x^2-3x+2) / (x^3(x - 2))
= A / x + B / x^2 + C / x^3 + D / (x - 2)
To find the values of A, B, C, and D, we can multiply both sides of the equation by the common denominator and equating the numerators:
ln(-x^3+9x^2-3x+2) = A(x^2)(x - 2) + B(x)(x - 2) + C(x - 2) + D(x^3)
To determine A, let's consider the limit as x approaches 0:
ln(2) = -2A
Therefore, A = -ln(2)/2
To determine B, let's consider the limit as x approaches 0 again:
ln(2) = -2B
Therefore, B = -ln(2)/2
To determine C, let's consider the limit as x approaches 0 a third time:
ln(2) = -2C
Therefore, C = -ln(2)/2
To determine D, let's consider the limit as x approaches 2 this time:
ln(2) = 4D
Therefore, D = ln(2)/4
Now that we have the values of A, B, C, and D, we can rewrite the rational function:
ln(-x^3+9x^2-3x+2) / (x^4-2x^3) = (-ln(2)/2) / x + (-ln(2)/2) / x^2 + (-ln(2)/2) / x^3 + (ln(2)/4) / (x - 2)
Now, we can integrate each term separately:
∫ (-ln(2)/2) / x dx = -ln(2)/2 * ln(x) + C1
∫ (-ln(2)/2) / x^2 dx = -ln(2)/2 * (-1/x) + C2 = ln(2)/2x + C2
∫ (-ln(2)/2) / x^3 dx = -ln(2)/2 * (-1/(2x^2)) + C3 = ln(2)/4x^2 + C3
∫ (ln(2)/4) / (x - 2) dx = (ln(2)/4) * ln|(x - 2)| + C4
Therefore, the indefinite integral of the given rational function is:
-ln(2)/2 * ln(x) + ln(2)/2x + ln(2)/4x^2 + (ln(2)/4) * ln|(x - 2)| + C
It appears that your approach was correct, and you should have arrived at this answer. Double-check your calculations and make sure you didn't make any calculation errors.