Q.(a) Find: ∫x−2x+2dx.
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🔒 Start your 14-day free trial to unlock the full solution →Part (a)Concept understanding — Rationalizing Denominator
Rationalizing the Denominator
Rationalizing the denominator means rewriting a fraction so that no radical (square root, cube root, …) is left on the bottom. It is algebraic housekeeping — the fraction's value never changes, because you only ever multiply by a cleverly disguised form of 1.
Why bother? A quotient like 21 is awkward to estimate (1÷1.414), but the equal form 22 is easy (1.414÷2≈0.707). Cleaner denominators are also easier to add, compare and simplify, and most answer keys expect this final form.
Case 1 — a single square root
Multiply top and bottom by that root:
53×55=535,
because 5×5=5 is rational. In general ba=bab.
Case 2 — a sum or difference with a root
Here multiplying by the root alone fails; use the conjugate, which turns the denominator into a difference of squares:
3+72×3−73−7=32−(7)22(3−7)=22(3−7)=3−7.
For b+ca, multiply by b−cb−c; the denominator becomes b2−c, a rational number.
Multiply both the numerator and the denominator by the same expression. Changing only the bottom changes the value of the fraction. …
Part (b)Concept understanding — Partial Fraction Decomposition
Partial Fraction Decomposition
Adding x−12+x+23 over a common denominator gives x2+x−25x+1. 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: ∫x−12dx=2log∣x−1∣ is immediate, while the combined fraction is not.
When it applies
You need a proper rational function, degP<degQ. 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 Q(x)P(x) with Q fully factored, each factor contributes a term:
- Distinct linear (ax+b) → ax+bA.
- Repeated linear (ax+b)n → ax+bA1+(ax+b)2A2+⋯+(ax+b)nAn.
- Irreducible quadratic (ax2+bx+c) → ax2+bx+cAx+B — 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 x2−3x+23x+5. Factor the denominator: (x−1)(x−2). Set
(x−1)(x−2)3x+5=x−1A+x−2B.
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
x2−3x+23x+5=x−1−8+x−211.
With distinct linear factors, substitute each factor's root to knock out all but one term — much faster than equating coefficients. …
Part (a)
Rationalise: x−2x+2=(x+2)(x−2)x+2=x2−4x+2. Split:
∫x2−4x+2dx=∫x2−4xdx+2∫x2−4dx=x2−4+2lnx+x2−4+C. …
- ∫x−2x+2dx=x2−4+2lnx+x2−4+C.
- ∫(x2+9)(x2+16)x2dx=74tan−14x−73tan−13x+C.
Part (a)
The square root of a ratio is awkward, so rationalise by multiplying inside by (x+2):
x−2x+2=x−2x+2⋅x+2x+2=(x+2)(x−2)x+2=x2−4x+2.
Now split the numerator:
∫x2−4x+2dx=I1∫x2−4xdx+I22∫x2−4dx.
I1: substitute u=x2−4, du=2xdx:
I1=21∫u−1/2du=u=x2−4.
I2: standard result ∫x2−a2dx=lnx+x2−a2 with a=2:
I2=2lnx+x2−4.
Adding, …
- CBSE 2026Set A1 markMCQQ.∫x2−a2dx=(a) a1tan−1ax+k(b) 2a1logx+ax−a+k(c) 2a1loga−xa+x+k(d) a1logx+ax−a+k
›Reveal solutionSolution
Standard integral: ∫x2−a2dx=2a1logx+ax−a+k.
Using partial fractions, x2−a21=(x−a)(x+a)1=2a1(x−a1−x+a1).
…
- CBSE 2024Set D1 markMCQQ.∫x(x+2)dx=(a) logx+2x+c(b) 21logx+2x+c(c) log∣x∣+c(d) log∣x+2∣+c
›Reveal solutionSolution
Partial fractions give 21logx+2x+c.
Write x(x+2)1=21(x1−x+21).
…
- CBSE 2022Set ANNUAL1 markMCQQ.∫x2−1dx=(a) sin−1x+k(b) 21logx+1x−1+k(c) 21logx−1x+1+k(d) 1−x2+k
›Reveal solutionSolution
∫x2−1dx=21logx+1x−1+k.
The standard result is ∫x2−a2dx=2a1logx+ax−a+k.
…
- CBSE 2021Set I1 markMCQQ.∫secx+tanxsecxdx=(a) tanx+secx+k(b) tanx−secx+k(c) secx+k(d) tanx+k
›Reveal solutionSolution
∫secx+tanxsecxdx=tanx−secx+k.
Multiply numerator and denominator by (secx−tanx). Since sec2x−tan2x=1: …
- CBSE 2021Set ANNUAL1 markMCQQ.∫(x−1)(x−2)xdx is equal to(a) logx−2(x−1)2+C(b) logx−1(x−2)2+C(c) log(x−2x−1)2+C(d) log∣(x−1)(x−2)∣+C
›Reveal solutionSolution
Partial fractions give (x−1)(x−2)x=x−1−1+x−22, integrating to logx−1(x−2)2+C.
Let (x−1)(x−2)x=x−1A+x−2B
x=A(x−2)+B(x−1)
At x=1: 1=−A⇒A=−1
At x=2: 2=B⇒B=2
…
- CBSE 2019Set ANNUAL1 markMCQQ.If 1/(x(x−3)) = A/x + B/(x−3), then the value of B is:(a) 1/2(b) −1/3(c) 1/3(d) −1/2
›Reveal solutionSolution
Clear the denominator and compare coefficients (or plug in x=3) to isolate B.
Write x(x−3)1=xA+x−3B.
Multiplying both sides by x(x−3):
1=A(x−3)+Bx
…
- CBSE 2019Set ANNUAL1 markMCQQ.The value of ∫ dx/(x²−a²) is:(a) (1/a) tan⁻¹(x/a) + C(b) (1/2a) log((x−a)/(x+a)) + C(c) sin⁻¹(x/a) + C(d) (1/2a) log((x+a)/(x−a)) + C
›Reveal solutionSolution
This is a standard result obtained by partial fractions: x2−a21=2a1(x−a1−x+a1).
∫x2−a2dx=2a1∫(x−a1−x+a1)dx=2a1[log∣x−a∣−log∣x+a∣]+C
…
- CBSE 2018Set ANNUAL1 markMCQQ.If (1+sinx)(2+sinx)1=(1+sinx)a+(2+sinx)b then a+b=(a) 0(b) 1(c) 2(d) 3
›Reveal solutionSolution
Partial fractions give a=1, b=−1, hence a+b=0.
Let t=sinx. Then (1+t)(2+t)1=1+ta+2+tb, so 1=a(2+t)+b(1+t).
…
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