Evaluate the definite integral: $\int_{1}^{2} \frac{5x^2}{x^2+4x+3} \, dx$

Evaluate the definite integral: ∫₁² (5x²)/(x²+4x+3) dx

Exercise 7.8 · Q16

\int_{1}^{2} \frac{5x^2}{x^2+4x+3} \, dx

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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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Divide (the fraction is improper), split the remainder by partial fractions, then integrate: the value is $5+\frac52\log\frac32-\frac{45}{2}\log\frac54\approx0.993$ .

Why divide first

The numerator $5x^2$ and denominator $x^2+4x+3$ have the same degree, so the fraction is improper. Partial fractions only work on a proper fraction, so polynomial division comes before anything else.

1. Polynomial division

Since $5(x^2+4x+3)=5x^2+20x+15$ ,

\frac{5x^2}{x^2+4x+3}=5-\frac{20x+15}{x^2+4x+3}.

2. Partial fractions on the remainder

Factor $x^2+4x+3=(x+1)(x+3)$ and write

\frac{20x+15}{(x+1)(x+3)}=\frac{A}{x+1}+\frac{B}{x+3}\ \Rightarrow\ 20x+15=A(x+3)+B(x+1).

Put $x=-1$ : $-5=2A\Rightarrow A=-\tfrac52$ . Put $x=-3$ : $-45=-2B\Rightarrow B=\tfrac{45}{2}$ .

Hence

$\frac{5x^2}{x^2+4x+3}=5-\left(\frac{-5/2}{x+1}+\frac{45/2}{x+3}\right)=5+\frac{5/2}{x+1}-\frac{45/2}{x+3}.$ …

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Q.Evaluate the definite integral: $\int_{1}^{2} \frac{5x^2}{x^2+4x+3} \, dx$

Concept understanding — Partial Fraction Decomposition

Partial Fraction Decomposition

When it applies

The standard forms

Worked example

Why divide first

1. Polynomial division

2. Partial fractions on the remainder

Method: Improper rational function — divide first, then partial fractions

Steps

Common Mistakes

Q.Evaluate the definite integral: ∫125x2x2+4x+3 dx\int_{1}^{2} \frac{5x^2}{x^2+4x+3} \, dx∫12​x2+4x+35x2​dx

Why divide first

1. Polynomial division

2. Partial fractions on the remainder

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Q.Evaluate the definite integral: $\int_{1}^{2} \frac{5x^2}{x^2+4x+3} \, dx$

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