Q.Find ∫x2−5x+6x2+1dx
Concept understanding — Polynomial Long Division
Polynomial Long Division
Dividing 137 by 4 asks "how many 4's fit into 137?" — answer 34, remainder 1. Polynomial long division is the same question with variables: how many times does the divisor fit into the dividend? You get a quotient polynomial plus a remainder whose degree is smaller than the divisor's. The only change from arithmetic is that you compare the highest power of the variable instead of place value.
For polynomials P(x) and D(x)=0 there are unique Q(x) and R(x) with
P(x)=D(x)Q(x)+R(x),degR<degD.
This is the Division Algorithm for Polynomials.
The routine
To divide P(x) by D(x), repeat until the remainder's degree drops below degD:
- Divide the leading term of the current dividend by the leading term of D(x) — this is the next quotient term.
- Multiply the whole divisor by that term.
- Subtract to get a new, lower-degree dividend, then repeat.
For example, dividing 2x3+3x2−5x+1 by x−2: the successive quotient terms are 2x2, then 7x, then 9, leaving remainder 19. So
2x3+3x2−5x+1=(x−2)(2x2+7x+9)+19.
The remainder 19 has degree 0<1, exactly as the algorithm requires.
Insert zero coefficients for missing terms — write x3+1 as x3+0x2+0x+1 — or the columns misalign during subtraction.
Why it matters
- If R(x)=0, then D(x) is a factor of P(x).
- Dividing by (x−a) leaves remainder P(a) — the Remainder Theorem (here P(2)=19).
- It reduces an improper rational function to a polynomial plus a proper fraction — the first step before partial fractions or integration.
Polynomial long division is introduced as early as the NCERT Class 9-10 Polynomials chapters and resurfaces as an essential prerequisite skill in the Class 12 Integrals chapter, wherever an improper rational function needs to be simplified before integration or partial fractions. Students searching 'polynomial long division examples class 10' or 'division algorithm for polynomials' will find this quotient-and-remainder method is exactly the same one tested in board exams at both levels.
Numerator and denominator have the same degree, so divide first.
Long division: x2−5x+6x2+1=1+x2−5x+65x−5.
Factor and split: x2−5x+6=(x−2)(x−3), and from 5x−5=A(x−3)+B(x−2), x=2⇒A=−5, x=3⇒B=10, so (x−2)(x−3)5x−5=x−2−5+x−310.
Integrate:
∫(1−x−25+x−310)dx=x−5log∣x−2∣+10log∣x−3∣+C.
x−5log∣x−2∣+10log∣x−3∣+C
Long division gives 1+(x−2)(x−3)5x−5; partial fractions then yield x−5log∣x−2∣+10log∣x−3∣+C.
Why divide first?
The fraction is improper — the numerator degree (2) equals the denominator degree (2). Partial fractions only apply to a proper fraction, so we first pull out the whole-number part by long division.
Step 1 — long division
x2 into x2 goes once. Subtract 1⋅(x2−5x+6) from x2+1:
(x2+1)−(x2−5x+6)=5x−5.
So
x2−5x+6x2+1=1+x2−5x+65x−5.
Step 2 — factor and decompose
x2−5x+6=(x−2)(x−3). Set
(x−2)(x−3)5x−5=x−2A+x−3B,5x−5=A(x−3)+B(x−2).
Put x=2: 5=A(−1)⇒A=−5. Put x=3: 10=B(1)⇒B=10.
Step 3 — integrate
∫(1−x−25+x−310)dx=x−5log∣x−2∣+10log∣x−3∣+C.
∫x2−5x+6x2+1dx=x−5log∣x−2∣+10log∣x−3∣+C
Method: Long Division First, Then Partial Fractions (Improper Rational Functions)
Use this when the numerator's degree is greater than or equal to the denominator's: divide before decomposing, because partial fractions only apply to proper fractions.
Steps
Step 1: Divide to separate the polynomial part.
Perform polynomial long division to write
D(x)N(x)=Q(x)+D(x)R(x),
where degR<degD. For x2−5x+6x2+1 this gives 1+x2−5x+65x−5.
Step 2: Decompose the proper remainder.
Factor the denominator and split D(x)R(x) into partial fractions, solving for the constants.
Step 3: Integrate every piece.
The quotient Q(x) integrates by the power rule; each partial fraction integrates to a logarithm. Combine and add C.
Common Mistakes
Mistake 1: Jumping straight to partial fractions.
Why it's wrong: the fraction is improper (deg numerator =deg denominator), so decomposition is invalid until you divide. Correct approach: do long division first.
Mistake 2: Stopping division too early or too late.
Why it's wrong: the remainder must have degree strictly less than the denominator's; otherwise the split is wrong. Correct approach: divide until degR<degD, giving remainder 5x−5 here.
Mistake 3: Forgetting to integrate the quotient term.
Why it's wrong: dropping the "1" from 1+⋯5x−5 loses the x term in the answer. Correct approach: integrate both the quotient and the partial fractions.
- CBSE 2024Set D1 markMCQQ.∫x2+1x4+1dx=(a) 3x3+c(b) 3x3−x+2tan−1x+c(c) 2tan−1x+c(d) 3x3+x+2tan−1x+c
›Reveal solutionSolution
Split the improper fraction by polynomial division: x2+1x4+1=x2−1+x2+12.
Write x4+1=(x4−1)+2=(x2−1)(x2+1)+2. Dividing by x2+1:
x2+1x4+1=x2−1+x2+12.
Integrate term by term:
∫(x2−1+x2+12)dx=3x3−x+2tan−1x+c.
✓Final answer(B) 3x3−x+2tan−1x+c
- CBSE 2022Set ANNUAL1 markMCQQ.∫x−1x2−1dx=(a) x+k(b) 2x2+x+k(c) 3x3−2x2+k(d) 2x2+2x+k
›Reveal solutionSolution
Simplify x−1x2−1=x+1, then integrate.
x−1x2−1=x−1(x−1)(x+1)=x+1 (for xeq1).
∫(x+1)dx=2x2+x+k.
✓Final answer(b) 2x2+x+k.
- CBSE 2021Set I1 markMCQQ.∫x2+1x4−1dx=(a) 3x3+2x+k(b) 3x3−2x+k(c) 3x3+x+k(d) 3x3−x+k
›Reveal solutionSolution
∫x2+1x4−1dx=3x3−x+k.
Factor the numerator as a difference of squares: x4−1=(x2−1)(x2+1). So
x2+1x4−1=x2+1(x2−1)(x2+1)=x2−1.
Integrate:
∫(x2−1)dx=3x3−x+k.
✓Final answer(d) 3x3−x+k.
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