Question

problem with integration:

http://gyazo.com/89f6c648f3cf2cf0b074b3c5ed419752

I know you're supposed to separate into partial fractions, but I get 5, -2 and 3 as the top numbers ABand C. could somebody walk me through just the separation of partial fractions?

Answer 1

First, A = 5 , B= -2, C = 1 [not 3]
Then,\[\int\limits(3x^2+9x+5)/(x(x+1)^2)dx= \int\limits (5/x) -( 2/(x+1)) +( 1/(x+1)^2) dx\]Now, you can use the sum and difference rule of the integration\[\int\limits\limits(3x^2+9x+5)/(x(x+1)^2)= \int\limits\limits (5/x) dx -\int\limits( 2/(x+1))dx +\int\limits( 1/(x+1)^2)dx\] Next, you can use the substitution to find the antiderivatives \[\int\limits\limits\limits(3x^2+9x+5)/(x(x+1)^2) =5\ln(x) - 2\ln(x+1) + (-1/x+1)\]

Answer 2

\[\int\limits_{}^{} (A/x) + \int\limits_{}^{} (B/(x+1) + \int\limits_{}^{} (C/(x+1)^2)\]
w/c leads to A=5, B=-2 and C=1
= 5ln x - 2 ln(x+1) - (1/(x+1)) ans....