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$\displaystyle\int \frac{(x+4)}{(x+1)^{2}}dx$ I broke it up into partial fractions:

$\frac{A}{(x+1)} + \frac{B}{(x+1)^{2}}$

Then I got confused by the common denominator aspect, wouldn’t B’s $(x+1)^{2} $cancel out completely the $(x+1)^{2}$ with its denominator so then it would look like:

$x + 4 = A(x+1) + B $?

Please help not sure how to solve

Thank you

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    $\begingroup$ Yes you got it. So A=1 and A+B = 4 $\endgroup$ – B.A Mar 31 '17 at 1:20
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Yeah that's fine. $x+4=Ax+A+B$ so $A=1$ and $A+B=4$ so $B=3$.

Then your integral becomes:

$$\int \frac{1}{x+1}+\frac{3}{(x+1)^2}\ dx$$

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You're almost there. Your last equation becomes

$x+4 = Ax + (A+B)$.

By comparing coefficients you can see that $A=1$ and $A+B=4$. Can you finish it now?

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