Evaluate: $\displaystyle\int\dfrac{x^5+x^4+x^3+x^2+4x+1}{x^3+1}\,dx$.

Question

Accepted Answer

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> [!TLDR]
> Polynomial long division reduces the integrand to a polynomial plus $\frac{3x}{x^3+1}$; partial fractions on the remainder (using $x^3+1=(x+1)(x^2-x+1)$) finish the integration.
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> **Step 1 — divide:** $x^5+x^4+x^3+x^2+4x+1$ by $x^3+1$:
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> $x^5+x^4+x^3+x^2+4x+1=(x^3+1)(x^2+x+1)+3x$
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> (quotient $x^2+x+1$, remainder $3x$ — verified: $(x^3+1)(x^2+x+1)=x^5+x^4+x^3+x^2+x+1$, and adding $3x$ gives $x^5+x^4+x^3+x^2+4x+1$ ✓).
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> So the integrand $=x^2+x+1+\dfrac{3x}{x^3+1}$.
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> **Step 2 — integrate the polynomial part:**
> $$\int(x^2+x+1)\,dx=\dfrac{x^3}{3}+\dfrac{x^2}{2}+x.$$
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> **Step 3 — partial fractions for $\dfrac{3x}{x^3+1}=\dfrac{3x}{(x+1)(x^2-x+1)}$:**
> $$\dfrac{3x}{(x+1)(x^2-x+1)}=\dfrac{A}{x+1}+\dfrac{Bx+C}{x^2-x+1}.$$
> $3x=A(x^2-x+1)+(Bx+C)(x+1)$. Put $x=-1$: $-3=3A\Rightarrow A=-1$. Comparing $x^2$: $0=A+B\Rightarrow B=1$. Comparing constants: $0=A+C\Rightarrow C=1$.
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> So $\dfrac{3x}{x^3+1}=\dfrac{-1}{x+1}+\dfrac{x+1}{x^2-x+1}$.
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> $$\int\dfrac{-1}{x+1}dx=-\ln|x+1|.$$
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> For $\displaystyle\int\dfrac{x+1}{x^2-x+1}dx$, write $x+1=\dfrac12(2x-1)+\dfrac32$: …

Q.Evaluate: $\displaystyle\int\dfrac{x^5+x^4+x^3+x^2+4x+1}{x^3+1}\,dx$ .

Concept understanding — Partial Fraction Decomposition

Partial Fraction Decomposition

When it applies

The standard forms

Worked example

Q.Evaluate: ∫x5+x4+x3+x2+4x+1x3+1 dx\displaystyle\int\dfrac{x^5+x^4+x^3+x^2+4x+1}{x^3+1}\,dx∫x3+1x5+x4+x3+x2+4x+1​dx.

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Q.Evaluate: $\displaystyle\int\dfrac{x^5+x^4+x^3+x^2+4x+1}{x^3+1}\,dx$ .

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