How do you integrate x^3/(x^2+2x+1) using partial fractions?

Apr 27, 2017

$\int {x}^{3} / \left({x}^{2} + 2 x + 1\right) \mathrm{dx} = {x}^{2} - 2 x + 3 \ln | x + 1 | + \frac{1}{x + 1}$

Explanation:

${x}^{3} / \left({x}^{2} + 2 x + 1\right) = \frac{x \left({x}^{2} + 2 x + 1\right) - 2 \left({x}^{2} + 2 x + 1\right) + 3 x + 2}{{x}^{2} + 2 x + 1}$

= $x - 2 + \frac{3 x + 2}{{x}^{2} + 2 x + 1}$. Let

$\frac{3 x + 2}{{x}^{2} + 2 x + 1} = \frac{3 x + 2}{x + 1} ^ 2 \equiv \frac{A}{x + 1} + \frac{B}{x + 1} ^ 2$

i.e. $3 x + 2 = A \left(x + 1\right) + B$

Now if $x = 0$, we have $A + B = 2$ and if $x = - 1$, we have $B = - 1$ and hence $A = 3$

Therefore $\frac{3 x + 2}{{x}^{2} + 2 x + 1} = \frac{3 x + 2}{x + 1} ^ 2 \equiv \frac{3}{x + 1} - \frac{1}{x + 1} ^ 2$

and ${x}^{3} / \left({x}^{2} + 2 x + 1\right) = x - 2 + \frac{3}{x + 1} - \frac{1}{x + 1} ^ 2$

hence $\int {x}^{3} / \left({x}^{2} + 2 x + 1\right) \mathrm{dx} = \int \left(x - 2 + \frac{3}{x + 1} - \frac{1}{x + 1} ^ 2\right) \mathrm{dx}$

= ${x}^{2} - 2 x + 3 \ln | x + 1 | + \frac{1}{x + 1}$

Apr 27, 2017

${\left(x + 1\right)}^{2} / 2 - 3 \left(x + 1\right) + 3 \ln | x + 1 | + \frac{1}{x + 1} + C .$

Explanation:

Though the Problem is requred to be solved using Partial Fractions,

as a Second Method, we show that the same can be solved

easily without its use.

Let, $I = \int {x}^{3} / \left({x}^{2} + 2 x + 1\right) \mathrm{dx} = \int {x}^{3} / {\left(x + 1\right)}^{2} \mathrm{dx} .$

Sbst.ing, $x + 1 = t , \text{ so that, } \mathrm{dx} = \mathrm{dt} , \mathmr{and} , x = t - 1.$

$\therefore I = \int {\left(t - 1\right)}^{3} / {t}^{2} \mathrm{dt} ,$

$= \int \frac{{t}^{3} - 1 - 3 t \left(t - 1\right)}{t} ^ 2 \mathrm{dt} ,$

$= \int \frac{{t}^{3} - 3 {t}^{2} + 3 t - 1}{t} ^ 2 \mathrm{dt} ,$

$= \int \left\{{t}^{3} / {t}^{2} - 3 {t}^{2} / {t}^{2} + 3 \frac{t}{t} ^ 2 - \frac{1}{t} ^ 2\right\} \mathrm{dt} ,$

$= \int \left\{t - 3 + \frac{3}{t} - \frac{1}{t} ^ 2\right\} \mathrm{dt} ,$

$= {t}^{2} / 2 - 3 t + 3 \ln | t | + \frac{1}{t} ,$

$\Rightarrow I = {\left(x + 1\right)}^{2} / 2 - 3 \left(x + 1\right) + 3 \ln | x + 1 | + \frac{1}{x + 1} + C .$

Enjoy Maths.!

N.B.: $I = {x}^{2} / 2 + x + \frac{1}{2} - 3 x - 3 + 3 \ln | x + 1 | + \frac{1}{x + 1} + C ,$

$= {x}^{2} / 2 - 2 x - \frac{5}{2} + 3 \ln | x + 1 | + \frac{1}{x + 1} + C ,$

$= {x}^{2} / 2 - 2 x + 3 \ln | x + 1 | + \frac{1}{x + 1} + c , w h e r e , c = C - \frac{5}{2} ,$ as

Respected Shwetank Mauria has derived!