## ggrree Group Title 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? 2 years ago 2 years ago

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1. myname

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 $\int\limits\limits\limits(3x^2+9x+5)/(x(x+1)^2) =5\ln(x) - 2\ln(x+1) + (-1/x+1)$

2. mark_o.

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