Integrate the following function: $\frac{3x + 5}{x^3 - x^2 - x + 1}$

Integrate the following function: (3x + 5)/(x³ - x – Class 12 Mathematics

Exercise 7.5 · Q9

\frac{3x + 5}{x^3 - x^2 - x + 1}

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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 denominator factors as $(x-1)^2(x+1)$ ; partial fractions give $\frac{-1/2}{x-1} + \frac{4}{(x-1)^2} + \frac{1/2}{x+1}$ , integrating to $-\frac12\log|x-1| - \frac{4}{x-1} + \frac12\log|x+1| + C$ .

Step 1 — factor the cubic

Group:

x^3-x^2-x+1 = x^2(x-1) - (x-1) = (x-1)(x^2-1) = (x-1)^2(x+1).

The repeated factor $(x-1)^2$ shapes the decomposition.

Step 2 — set up the form

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

Step 3 — solve

Clearing denominators: $3x+5 = A(x-1)(x+1) + B(x+1) + C(x-1)^2.$

$x=1$ : $\ 8 = B(2) \Rightarrow B = 4.$
$x=-1$ : $\ 2 = C(-2)^2 = 4C \Rightarrow C = \tfrac12.$
Coefficient of $x^2$ : $\ 0 = A + C \Rightarrow A = -\tfrac12.$

Step 4 — integrate term by term

$\displaystyle \int \frac{-1/2}{x-1}\,dx = -\frac12\log|x-1|.$ …

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Q.Integrate the following function: $\frac{3x + 5}{x^3 - x^2 - x + 1}$

Concept understanding — Partial Fraction Decomposition

Partial Fraction Decomposition

When it applies

The standard forms

Worked example

Step 1 — factor the cubic

Step 2 — set up the form

Step 3 — solve

Step 4 — integrate term by term

Method: Factor by Grouping, then Partial Fractions

Steps

Common Mistakes

Q.Integrate the following function: 3x+5x3−x2−x+1\frac{3x + 5}{x^3 - x^2 - x + 1}x3−x2−x+13x+5​

Step 1 — factor the cubic

Step 2 — set up the form

Step 3 — solve

Step 4 — integrate term by term

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Q.Integrate the following function: $\frac{3x + 5}{x^3 - x^2 - x + 1}$

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