Integrate the following function: $\frac{5x}{(x + 1)(x^2 - 4)}$

Integrate the following function: (5x)/((x + 1)(x² – Class 12 Mathematics

Exercise 7.5 · Q11

\frac{5x}{(x + 1)(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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Factor the denominator into three distinct linear factors, decompose into partial fractions, and integrate each piece to a logarithm: $\dfrac{5}{3}\log|x+1|+\dfrac{5}{6}\log|x-2|-\dfrac{5}{2}\log|x+2|+C$ .

Why partial fractions

A single fraction like $\dfrac{5x}{(x+1)(x^2-4)}$ has no direct antiderivative, but a sum of pieces of the form $\dfrac{A}{x-a}$ does — each integrates to $A\log|x-a|$ . The whole job is to rewrite the fraction as such a sum.

Step 1 — Factor the denominator completely

x^2-4=(x-2)(x+2),\qquad\text{so}\qquad (x+1)(x^2-4)=(x+1)(x-2)(x+2).

Three distinct linear factors, and the numerator degree $1$ is less than the denominator degree $3$ , so the fraction is proper and we can decompose directly.

Step 2 — Set up the decomposition

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

Step 3 — Clear denominators and solve

5x=A(x-2)(x+2)+B(x+1)(x+2)+C(x+1)(x-2).

Substitute each root so two terms vanish:

$x=-1$ : $\;5(-1)=A(-3)(1)\Rightarrow -5=-3A\Rightarrow A=\dfrac{5}{3}$
$x=2$ : $\;5(2)=B(3)(4)\Rightarrow 10=12B\Rightarrow B=\dfrac{5}{6}$ …

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Q.Integrate the following function: $\frac{5x}{(x + 1)(x^2 - 4)}$

Concept understanding — Partial Fraction Decomposition

Partial Fraction Decomposition

When it applies

The standard forms

Worked example

Why partial fractions

Step 1 — Factor the denominator completely

Step 2 — Set up the decomposition

Step 3 — Clear denominators and solve

Method: Partial Fractions — Distinct Linear Factors

Steps

Common Mistakes

Q.Integrate the following function: 5x(x+1)(x2−4)\frac{5x}{(x + 1)(x^2 - 4)}(x+1)(x2−4)5x​

Why partial fractions

Step 1 — Factor the denominator completely

Step 2 — Set up the decomposition

Step 3 — Clear denominators and solve

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Q.Integrate the following function: $\frac{5x}{(x + 1)(x^2 - 4)}$

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