Question

How do you integrate $\int {\dfrac{{3{x^2} - 2x + 5}}{{{{({x^2} + 1)}^2}}}}$ using partial fractions?

Answer 1

In this question, we will try to represent the complex expression in the simplified manner in the combination of fractions which has the same value and then we will integrate the partial fractions separately to get the required solution.

Complete step-by-step solution:
We have the expression as:
$\int {\dfrac{{3{x^2} - 2x + 5}}{{{{({x^2} + 1)}^2}}}}$
Now since the expression has a squared element, the skeleton of writing the expression in the form of partial fractions will be:
$\Rightarrow$ $\dfrac{{3{x^2} - 2x + 5}}{{{{({x^2} + 1)}^2}}} = \dfrac{{A + Bx}}{{{x^2} + 1}} + \dfrac{{C + Dx}}{{{{({x^2} + 1)}^2}}}$
Now on taking the lowest-common multiple on both the sides and cancelling the terms, we get:
$\Rightarrow$ $3{x^2} - 2x + 5 = (A + Bx)({x^2} + 1) + C + Dx$
Now on multiplying the terms in the right-hand side, we get:
$\Rightarrow$ $3{x^2} - 2x + 5 = A + A{x^2} + B{x^3} + 1 + Bx + Dx + C$
Now on grouping the terms we get:
$\Rightarrow$ $3{x^2} - 2x + 5 = A + A{x^2} + B{x^3} + 1 + (B + D)x + C$
Now on comparing the $x$ terms in the left-hand side and the right-hand side, we get:
$A + C = 5 \to (1)$
$B + D = - 2 \to (2)$
$A = 3 \to (3)$
$B = 0 \to (4)$
On substituting the value of equation $(4)$ into equation $(2)$ , we get:
$0 + D = - 2$
Therefore, we get:
$D = - 2$
On substituting the value of equation $(3)$ into equation $(1)$ , we get:
$3 + C = 5$
On rearranging and simplifying, we get:
$C = 2$
Therefore, we can write the expression in the form of partial fractions as:
$\Rightarrow$ $\dfrac{{3{x^2} - 2x + 5}}{{{{({x^2} + 1)}^2}}} = \dfrac{{3 + 0x}}{{{x^2} + 1}} + \dfrac{{2 - 2x}}{{{{({x^2} + 1)}^2}}}$
On simplifying, we get:
$\Rightarrow$ $\dfrac{{3{x^2} - 2x + 5}}{{{{({x^2} + 1)}^2}}} = \dfrac{3}{{{x^2} + 1}} + \dfrac{{2 - 2x}}{{{{({x^2} + 1)}^2}}}$
Therefore, the integration can be written as:
$\Rightarrow$ $\int {\dfrac{3}{{{x^2} + 1}} + \dfrac{{2 - 2x}}{{{{({x^2} + 1)}^2}}}} dx$
On splitting the integration, we get:
\Rightarrow$$$\int {\dfrac{3}{{{x^2} + 1}}dx + \int {\dfrac{{2 - 2x}}{{{{({x^2} + 1)}^2}}}} } dx$$ The first part can be solved using the formula \int {\dfrac{1}{{1 + {x^2}}} = {{\tan }^{ - 1}}x + c} \to (5)and the second part can be solved using the parametric substitution as $$\int {\dfrac{{2 - 2x}}{{{{({x^2} + 1)}^2}}}} = \dfrac{{{{\tan }^{ - 1}}x}}{2} + \dfrac{1}{2} \times \dfrac{{x + 1}}{{{x^2} + 1}} + c \to (6)$$ therefore on combining both the equations, we can write the solution as: \Rightarrow$$3{\tan ^{ - 1}}x + {\tan ^{ - 1}}x + \dfrac{{x + 1}}{{{x^2} + 1}} + c$, which can be simplified as:$\Rightarrow$$4{\tan ^{ - 1}}x + \dfrac{{x + 1}}{{{x^2} + 1}} + c$

$4{\tan ^{ - 1}}x + \dfrac{{x + 1}}{{{x^2} + 1}} + c$ is the answer for the given question.

Note: It is to be noted that integration and derivatives are the reverse of each other. If the derivative of a term $a$ is $b$ , then the integration of the term $b$ will be $a$ .
And unlike derivatives, there is no chain rule to simplify the expression directly therefore, one of the simplification measures which is partial fraction is used here.
Partial fraction is a method which involves expressing a fraction as a sum of two or more polynomials or fractions, for the purpose of simplification of the term to be integrated.