Integrate ( (1/ (x+3)(x^2+8x+15))dx with NO Bounds I got that the integral above

Question

Integrate ( (1/ (x+3)(x^2+8x+15))dx with NO Bounds

I got that the integral above can also equate to

=(integral) [ 1/(x+3)^2(x+5)

and by using partial fractions It would end up looking like

=[ (a/ x+3) + (b/(x+3)^2) + (c/(x+5)) ]

I know after that you should cancel out denominators by multiplying the denominator of (x+3)^2(x+5) to both sides. When I did this I got

1= A(x+3)(x+5) + B(x+5) + C(x+3)^2

=(A+C)x^2 + (8A+B+6C)x + 15A + 5B +9C

Yet when I try to Equate the Coeff. I Cannot seem to get any of the numbers to add up.

0=A +C therefore C= -A

0=8A + B+ 6C

1= 15A + 5B+ 9C

By assuming that C= -A Then the last two equations would give

0= 2A +B

1= 6A +5B

But I cannot get those equations to cancel out to solve for one variable. Therefore, I am assuming that I did somethign wrong (most likely during the cancellation of denominators). If so, can you tell me where I did so and the correct solution?

Explanation / Answer

by solving the equation 2A +B =0 and 6A + 5B =1,

since B= -2A

so 6A - 10A =1

A= -1/4, and B =1/2 and C= -A = 1/4

so the integral of (1/(x+3)^2 (x+5))

= integral of ( -1/ (4(x+3)) + 1/ 2((x+3)^2) + 1/ (4(x+5)) )

= (-1/4) ln(x+3) -1/ (2(x+3)) +(1/4) ln(x+5)

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