How do you integrate $int ( (dx) / ( x(x+1)^2 ) )$ using partial fractions?

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Answer 1 · 2016-02-08T03:23:15

$int frac{1}{x(x+1)^2} dx = ln|x| - ln|x+1| + 1/(x+1) + "Constant"$

As stated by the question, we must first express $frac{1}{x(x+1)^2}$ in partial fractions, before we perform the integration.

$frac{1}{x(x+1)^2} -= A/x + B/(x+1) + C/(x+1)^2$ ,

where $A$ , $B$ and $C$ are constants to be determined, such that the equality holds true for all possible values of $x$ .

Which means,

$1 -= A(x+1)^2 + Bx(x+1) + Cx$ .

Simplifying gives

$(A+B)*x^2 + (2A + B + C)*x + A -= 1$

We compare the coefficients (to zero) to get 3 simultaneous linear equations,

$A+B=0$
$2A+B+C=0$
$A=1$

The system of equations is solved easily to yield

$A=1$
$B=-1$
$C=-1$

Therefore,

$frac{1}{x(x+1)^2} -= 1/x - 1/(x+1) - 1/(x+1)^2$ .

Now, to integrate.

$int frac{1}{x(x+1)^2} dx = int (1/x - 1/(x+1) - 1/(x+1)^2) dx$

$= ln|x| - ln|x+1| + 1/(x+1) + "Constant"$

How do you integrate int ( (dx) / ( x(x+1)^2 ) )int ( (dx) / ( x(x+1)^2 ) ) using partial fractions?