Integrate the following function: $\frac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}$

Integrate the following function: ((x²+1)(x²+2))/((x²+3)(x²+4))

Exercise 7.5 · Q18

\frac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}

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Concept understanding — Partial Fraction Decomposition

Partial Fraction Decomposition

Adding $\frac{2}{x-1}+\frac{3}{x+2}$ over a common denominator gives $\frac{5x+1}{x^2+x-2}$ . Partial fraction decomposition reverses this — it breaks one complicated rational function back into a sum of simple pieces. Those pieces are far easier to integrate: $\int \frac{2}{x-1}\,dx = 2\log|x-1|$ is immediate, while the combined fraction is not.

When it applies

You need a proper rational function, $\deg P < \deg Q$ . If the numerator's degree is equal or higher, first do polynomial long division. The denominator $Q(x)$ must be factored into linear and/or irreducible quadratic factors.

The standard forms

For $\frac{P(x)}{Q(x)}$ with $Q$ fully factored, each factor contributes a term:

Distinct linear $(ax+b)$ $\to$ $\dfrac{A}{ax+b}$ .
Repeated linear $(ax+b)^n$ $\to$ $\dfrac{A_1}{ax+b}+\dfrac{A_2}{(ax+b)^2}+\cdots+\dfrac{A_n}{(ax+b)^n}$ .
Irreducible quadratic $(ax^2+bx+c)$ $\to$ $\dfrac{Ax+B}{ax^2+bx+c}$ — a linear numerator, not just a constant.

The unknown constants are found from the resulting equations; the decomposition is unique, which is what lets us solve for them systematically.

Worked example

Decompose $\frac{3x+5}{x^2-3x+2}$ . Factor the denominator: $(x-1)(x-2)$ . Set

\frac{3x+5}{(x-1)(x-2)} = \frac{A}{x-1}+\frac{B}{x-2}.

Clear denominators: $3x+5 = A(x-2)+B(x-1)$ . Substituting the roots, $x=1$ gives $8=-A$ so $A=-8$ , and $x=2$ gives $11=B$ . Hence

\frac{3x+5}{x^2-3x+2} = \frac{-8}{x-1}+\frac{11}{x-2}.

Tip

With distinct linear factors, substitute each factor's root to knock out all but one term — much faster than equating coefficients. …

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The fraction is improper (equal degrees), so it equals $1$ minus a proper fraction; decomposing in $x^2$ and integrating gives $x+\dfrac{2}{\sqrt3}\tan^{-1}\dfrac{x}{\sqrt3}-3\tan^{-1}\dfrac{x}{2}+C$ .

Step 1 — Reduce the improper fraction

Expand top and bottom:

(x^2+1)(x^2+2)=x^4+3x^2+2,\qquad (x^2+3)(x^2+4)=x^4+7x^2+12.

Since both are degree $4$ , divide. Subtracting the denominator from the numerator,

\frac{x^4+3x^2+2}{x^4+7x^2+12}=1+\frac{(x^4+3x^2+2)-(x^4+7x^2+12)}{x^4+7x^2+12}=1+\frac{-4x^2-10}{(x^2+3)(x^2+4)}.

Step 2 — Partial fractions in $x^2$

Both denominator factors are quadratics in $x$ but linear in $y=x^2$ , so substitute $y=x^2$ :

\frac{-4y-10}{(y+3)(y+4)}=\frac{a}{y+3}+\frac{b}{y+4},\qquad -4y-10=a(y+4)+b(y+3).

$y=-3$ : $\;-4(-3)-10=2=a(1)\Rightarrow a=2$
$y=-4$ : $\;-4(-4)-10=6=b(-1)\Rightarrow b=-6$

Restoring $y=x^2$ :

\frac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}=1+\frac{2}{x^2+3}-\frac{6}{x^2+4}.

Step 3 — Integrate …

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Q.Integrate the following function: $\frac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}$

Concept understanding — Partial Fraction Decomposition

Partial Fraction Decomposition

When it applies

The standard forms

Worked example

Step 1 — Reduce the improper fraction

Step 2 — Partial fractions in $x^2$

Step 3 — Integrate …

Method: Reduce an improper rational function, then decompose

Steps

Common Mistakes

Q.Integrate the following function: (x2+1)(x2+2)(x2+3)(x2+4)\frac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}(x2+3)(x2+4)(x2+1)(x2+2)​

Step 1 — Reduce the improper fraction

Step 2 — Partial fractions in x2x^2x2

Step 3 — Integrate …

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Q.Integrate the following function: $\frac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}$

Step 2 — Partial fractions in $x^2$

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