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

> [!TLDR] > Resolve the integrand into partial fractions, then integrate each piece with $\int\frac{dx}{x^2+a^2}=\frac{1}{a}\tan^{-1}\frac{x}{a}$. The result is $\frac{1}{3}\tan^{-1}x-\frac{1}{6}\tan^{-1}\frac{x}{2}+C$. ## Why partial fractions The denominator is a product of two irreducible quadratics. Splitting the fraction into pieces, each over a single quadratic, lets us apply the standard arctangent formula. > [!FORMULA] > $$\int \frac{dx}{x^2 + a^2} = \frac{1}{a}\tan^{-1}\frac{x}{a} + C$$ ## Step-by-step solution **1. Set up the decomposition.** Since the numerator is constant, treat $x^2$ as the variable: $$\frac{1}{(x^2+1)(x^2+4)} = \frac{A}{x^2+1} + \frac{B}{x^2+4} \quad\Rightarrow\quad 1 = A(x^2+4) + B(x^2+1).$$ **2. Solve for the constants.** Put $x^2=-1$: $\;1 = 3A \Rightarrow A=\tfrac{1}{3}$. Put $x^2=-4$: $\;1 = -3B \Rightarrow B=-\tfrac{1}{3}$. **3. Rewrite the integrand.** …

Integrate the function (1)/((x²+1)(x²+4))

Miscellaneous Exercise · Q14

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

Uttarakhand UbseTextbookSubjective· 3mImportance★★★★★

Appeared in past exams:TG EAPCET 2023· Set eng-2023-05-14-FN· 1mexact

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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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Resolve the integrand into partial fractions, then integrate each piece with $\int\frac{dx}{x^2+a^2}=\frac{1}{a}\tan^{-1}\frac{x}{a}$ . The result is $\frac{1}{3}\tan^{-1}x-\frac{1}{6}\tan^{-1}\frac{x}{2}+C$ .

Why partial fractions

The denominator is a product of two irreducible quadratics. Splitting the fraction into pieces, each over a single quadratic, lets us apply the standard arctangent formula.

\int \frac{dx}{x^2 + a^2} = \frac{1}{a}\tan^{-1}\frac{x}{a} + C

Step-by-step solution

1. Set up the decomposition. Since the numerator is constant, treat $x^2$ as the variable:

\frac{1}{(x^2+1)(x^2+4)} = \frac{A}{x^2+1} + \frac{B}{x^2+4} \quad\Rightarrow\quad 1 = A(x^2+4) + B(x^2+1).

2. Solve for the constants. Put $x^2=-1$ : $\;1 = 3A \Rightarrow A=\tfrac{1}{3}$ . Put $x^2=-4$ : $\;1 = -3B \Rightarrow B=-\tfrac{1}{3}$ .

3. Rewrite the integrand. …

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

Concept understanding — Partial Fraction Decomposition

Partial Fraction Decomposition

When it applies

The standard forms

Worked example

Why partial fractions

Step-by-step solution

Method: Partial fractions in $x^2$ for a product of irreducible quadratics

Steps

Common Mistakes

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

Why partial fractions

Step-by-step solution

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

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