Question

How do you integrate $\dfrac{{{x^3}}}{{{x^2} + 2x + 1}}$ using partial fraction?

Answer 1

Hint: In this question, we are given an expression and we have been asked to integrate it using partial fraction. But you will notice that the numerator is greater than the denominator. In such a situation, we will divide them both so that the numerator becomes smaller than the denominator. Then, we will see which partial fraction identity will be used here. After we know which formula will be used, we will apply that.

Complete step-by-step solution:
We have been given an expression $\dfrac{{{x^3}}}{{{x^2} + 2x + 1}}$ and we have been asked to integrate it using partial fractions. But if you observe carefully, the numerator of the given expression is greater than the denominator. So, we will divide them first in order to get a standard expression.
Dividing the numerator by the denominator, we get,
                              \[x-2\]
\[{x^2} + 2x + 1)\overline {\,x^3\,\,\,\,\,\,\,\,} \]
                \[
 \underline { (-)\,\,\, {x^3} + 2{x^2}} + x \\
    \,\,\,\, \,\,\,\,\,\,\,\, \, 0 - {2}{x^2} + x \\
 \]
                 \[
    (-) \,\,\,\, \,\,\,\,\, {-2}{x^2} -4x -2 \\
\overline {\,\,\,\, \,\,\,\,\, \,\,\,\, \,\,\,\,\, \,\,\,\,\, \,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\, {3x+2}} \\
 \]
Hence, the quotient is $x - 2$ and the remainder is $3x + 2$. Now, we will write the division in the form of–
$ \Rightarrow Q + \dfrac{R}{D}$
Putting all the values, we get,
$ \Rightarrow x - 2 + \dfrac{{3x + 2}}{{{x^2} + 2x + 1}}$
Therefore, $\dfrac{{{x^3}}}{{{x^2} + 2x + 1}}$$ = x - 2 + \dfrac{{3x + 2}}{{{x^2} + 2x + 1}}$
Now, we have to integrate the RHS of the equation shown above.
We will apply a partial fraction on $\dfrac{{3x + 2}}{{{x^2} + 2x + 1}}$. Let us find out the rule of partial fraction which will be used here. But first, we will have to find factors of the denominator.
$ \Rightarrow \dfrac{{3x + 2}}{{{x^2} + 2x + 1}} = \dfrac{{3x + 2}}{{{x^2} + x + x + 1}}$
 $ \Rightarrow \dfrac{{3x + 2}}{{{x^2} + 2x + 1}} = \dfrac{{3x + 2}}{{{{\left( {x + 1} \right)}^2}}}$
If we observe, $\dfrac{{3x + 2}}{{{{\left( {x + 1} \right)}^2}}}$ is similar to $\dfrac{{Ax + B}}{{{{\left( {x + a} \right)}^2}}}$. Now, we will use this rule and integrate the expression.
We know that,
$ \Rightarrow \dfrac{{Ax + B}}{{{{\left( {x + a} \right)}^2}}} = \dfrac{A}{{x + a}} + \dfrac{B}{{{{\left( {x + a} \right)}^2}}}$
Putting all the values,
$ \Rightarrow \dfrac{{3x + 2}}{{{{\left( {x + 1} \right)}^2}}} = \dfrac{A}{{x + 1}} + \dfrac{B}{{{{\left( {x + 1} \right)}^2}}}$
Taking LCM in the denominator,
$ \Rightarrow \dfrac{{3x + 2}}{{{{\left( {x + 1} \right)}^2}}} = \dfrac{{A\left( {x + 1} \right)}}{{{{\left( {x + 1} \right)}^2}}} + \dfrac{B}{{{{\left( {x + 1} \right)}^2}}}$
Solving to find the value of A and B,
$ \Rightarrow 3x + 2 = A\left( {x + 1} \right) + B$ …. (2)
In order to find the required values, we will assume certain values of x.
Let us put $x = - 1$.
$ \Rightarrow 3\left( { - 1} \right) + 2 = A\left( { - 1 + 1} \right) + B$
Simplifying,
$ \Rightarrow - 3 + 2 = B = - 1$
Now, we will put $x = 0$
$ \Rightarrow 2 = A\left( {0 + 1} \right) + B$
$ \Rightarrow 2 = A + B$
Putting $B = - 1$,
$ \Rightarrow 2 = A - 1$
Hence, $A = 3$.
Now, we get,
$ \Rightarrow \int {\left( {x - 2 + \dfrac{3}{{x + 1}} - \dfrac{1}{{{{\left( {x + 1} \right)}^2}}}} \right)dx} $

Integrating each term,
$ \Rightarrow \dfrac{{{x^2}}}{2} - 2x + 3\log \left| {x + 1} \right| + \dfrac{1}{{x + 1}} + C$

Note: In the last step, how did we integrate $\dfrac{{ - 1}}{{{{\left( {x + 1} \right)}^2}}}$?
Let $1 + x = t$. Differentiating both the sides, we get,
$ \Rightarrow dx = dt$
We have $ \Rightarrow \int {\dfrac{{ - 1}}{{{{\left( {x + 1} \right)}^2}}}} dx$
Putting in the equation,
$ \Rightarrow - \int {\dfrac{1}{{{t^2}}}} dt$
We can also write it as –
$ \Rightarrow - \int {{t^{ - 2}}dt} $
Using the formula, $\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} + C$
$ \Rightarrow - \dfrac{{{t^{ - 2 + 1}}}}{{ - 2 + 1}} + C$
$ \Rightarrow \dfrac{1}{t} + C$
Putting $1 + x = t$,$ \Rightarrow \dfrac{1}{{1 + x}} + C$