Find $\int \dfrac{x^2 + x + 1}{(x+2)(x^2+1)}\, dx$

Find ∫ (x² + x + 1)/((x+2)(x²+1)) dx

Worked Examples · Example 16

\int \dfrac{x^2 + x + 1}{(x+2)(x^2+1)}\, 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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Partial fractions give $A=\tfrac35,\,B=\tfrac25,\,C=\tfrac15$ , and the integral is $\dfrac35\log|x+2|+\dfrac15\log(x^2+1)+\dfrac15\tan^{-1}x+C$ .

Choosing the form

The denominator $(x+2)(x^2+1)$ has a linear factor $(x+2)$ — needing a constant numerator — and an irreducible quadratic $(x^2+1)$ — needing a linear numerator $Bx+C$ :

\frac{x^2+x+1}{(x+2)(x^2+1)}=\frac{A}{x+2}+\frac{Bx+C}{x^2+1}.

Solving for the constants

Multiply through by $(x+2)(x^2+1)$ :

x^2+x+1=A(x^2+1)+(Bx+C)(x+2)=(A+B)x^2+(2B+C)x+(A+2C).

Match coefficients:

A+B=1,\qquad 2B+C=1,\qquad A+2C=1.

From the first, $A=1-B$ ; substituting into the third gives $B=2C$ ; then $2(2C)+C=1\Rightarrow C=\tfrac15$ , so $B=\tfrac25$ and $A=\tfrac35$ .

Integrating

$\int\frac{3/5}{x+2}\,dx=\frac35\log|x+2|.$ …

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Q.Find $\int \dfrac{x^2 + x + 1}{(x+2)(x^2+1)}\, dx$

Concept understanding — Partial Fraction Decomposition

Partial Fraction Decomposition

When it applies

The standard forms

Worked example

Choosing the form

Solving for the constants

Integrating

Method: Partial Fractions with a Linear Factor and an Irreducible Quadratic

Steps

Common Mistakes

Q.Find ∫x2+x+1(x+2)(x2+1) dx\int \dfrac{x^2 + x + 1}{(x+2)(x^2+1)}\, dx∫(x+2)(x2+1)x2+x+1​dx

Choosing the form

Solving for the constants

Integrating

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Q.Find $\int \dfrac{x^2 + x + 1}{(x+2)(x^2+1)}\, dx$

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