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

Integrate the following function: (2x - 3)/((x² - – Class 12 Mathematics

Exercise 7.5 · Q10

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

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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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Three distinct linear factors give $A=-\tfrac{1}{10}$ , $B=\tfrac52$ , $C=-\tfrac{24}{5}$ ; integrating (with the $\tfrac12$ from $2x+3$ ) gives $-\frac{1}{10}\log|x-1| + \frac52\log|x+1| - \frac{12}{5}\log|2x+3| + C$ .

Step 1 — factor completely

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

Three distinct linear factors, numerator of lower degree — a clean partial-fraction case.

Step 2 — set up

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

Clearing: $2x-3 = A(x+1)(2x+3) + B(x-1)(2x+3) + C(x-1)(x+1).$

Step 3 — plug in the roots

$x=1$ : $\ 2(1)-3 = -1 = A(2)(5) = 10A \Rightarrow A = -\tfrac{1}{10}.$
$x=-1$ : $\ 2(-1)-3 = -5 = B(-2)(1) = -2B \Rightarrow B = \tfrac52.$
$x=-\tfrac32$ : $\ 2(-\tfrac32)-3 = -6 = C\left(-\tfrac32-1\right)\left(-\tfrac32+1\right) = C\left(-\tfrac52\right)\left(-\tfrac12\right) = \tfrac54 C \Rightarrow C = -\tfrac{24}{5}.$

Step 4 — integrate

$\int \frac{2x-3}{(x^2-1)(2x+3)}\,dx = -\frac{1}{10}\int\frac{dx}{x-1} + \frac52\int\frac{dx}{x+1} - \frac{24}{5}\int\frac{dx}{2x+3}.$ …

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

Concept understanding — Partial Fraction Decomposition

Partial Fraction Decomposition

When it applies

The standard forms

Worked example

Step 1 — factor completely

Step 2 — set up

Step 3 — plug in the roots

Step 4 — integrate

Method: Partial Fractions with a Non-Monic Linear Factor

Steps

Common Mistakes

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

Step 1 — factor completely

Step 2 — set up

Step 3 — plug in the roots

Step 4 — integrate

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

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