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

Nov 2, 2017

The partial fraction equation is:

$\frac{1}{x {\left(x + 1\right)}^{2}} = \frac{A}{x} + \frac{B}{x + 1} + \frac{C}{x + 1} ^ 2$

Multiply both sides by $x {\left(x + 1\right)}^{2}$

$1 = A {\left(x + 1\right)}^{2} + B x \left(x + 1\right) + C x$

Eliminate B and C by Letting $x = 0$:

$1 = A {\left(0 + 1\right)}^{2}$

$A = 1$

$1 = {\left(x + 1\right)}^{2} + B x \left(x + 1\right) + C x$

Eliminate B by letting $x = - 1$

$1 = C \left(- 1\right)$

$C = - 1$

$1 = {\left(x + 1\right)}^{2} + B x \left(x + 1\right) - x$

Let $x = 1$:

$1 = {\left(1 + 1\right)}^{2} + B \left(1\right) \left(1 + 1\right) - 1$

$2 - 4 = 2 B$

$B = - 1$

The partial fraction expansion is:

$\frac{1}{x {\left(x + 1\right)}^{2}} = \frac{1}{x} - \frac{1}{x + 1} - \frac{1}{x + 1} ^ 2$

Check:

1/x -1/(x+1)-1/(x+1)^2 = 1/x(x+1)^2/(x+1)^2-1/(x+1)(x(x+1))/(x(x+1)) - 1/(x+1)^2 x/x

$\frac{1}{x} - \frac{1}{x + 1} - \frac{1}{x + 1} ^ 2 = \frac{{x}^{2} + 2 x + 1 - {x}^{2} - x - x}{x {\left(x + 1\right)}^{2}}$

$\frac{1}{x} - \frac{1}{x + 1} - \frac{1}{x + 1} ^ 2 = \frac{1}{x {\left(x + 1\right)}^{2}}$

This checks.

The original integrand is equal to the partial fractions:

$\int \left(\frac{\mathrm{dx}}{x {\left(x + 1\right)}^{2}}\right) = \int \frac{1}{x} - \frac{1}{x + 1} - \frac{1}{x + 1} ^ 2 \mathrm{dx}$

Separate into 3 integrals:

$\int \left(\frac{\mathrm{dx}}{x {\left(x + 1\right)}^{2}}\right) = \int \frac{1}{x} \mathrm{dx} - \int \frac{1}{x + 1} \mathrm{dx} - \int \frac{1}{x + 1} ^ 2 \mathrm{dx}$

These integrals are well known:

$\int \left(\frac{\mathrm{dx}}{x {\left(x + 1\right)}^{2}}\right) = \ln | x | - \ln | x + 1 | + \frac{1}{x + 1} + C$