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Integral Calculus: JEE Main Mathematics Question with Solution Let
α ∈ ( 0 , 1 ) \alpha \in(0,1) α ∈ ( 0 , 1 ) and
β = log e ( 1 − α ) \beta=\log_e(1-\alpha) β = log e ( 1 − α ) . Let
P n ( x ) = x + x 2 2 + x 3 3 + ⋯ + x n n , x ∈ ( 0 , 1 ) P_n(x)=x+\frac{x^2}{2}+\frac{x^3}{3}+\dots+\frac{x^n}{n}, x \in(0,1) P n ( x ) = x + 2 x 2 + 3 x 3 + ⋯ + n x n , x ∈ ( 0 , 1 ) . Then the integral
∫ 0 α t 50 1 − t d t \int_0^\alpha \frac{t^{50}}{1-t}dt ∫ 0 α 1 − t t 50 d t is equal to
Step-by-step solution View Correct answer
Rewriting t 50 1 − t = t 50 − 1 1 − t + 1 1 − t = − ( 1 + t + ⋯ + t 49 ) + 1 1 − t \frac{t^{50}}{1-t} = \frac{t^{50}-1}{1-t} + \frac{1}{1-t} = -(1 + t + \dots + t^{49}) + \frac{1}{1-t} 1 − t t 50 = 1 − t t 50 − 1 + 1 − t 1 = − ( 1 + t + ⋯ + t 49 ) + 1 − t 1 − P 50 ( α ) − ln ( 1 − α ) = − ( β + P 50 ( α ) ) -P_{50}(\alpha) - \ln(1-\alpha) = -(\beta + P_{50}(\alpha)) − P 50 ( α ) − ln ( 1 − α ) = − ( β + P 50 ( α )) Option analysis
Why each option works or fails A · β + P 50 ( α ) \beta+P_{50}(\alpha) β + P 50 ( α ) Dropping the negative sign on both the polynomial terms and the logarithmic integral during integration. Observe that t 50 1 − t = − 1 − t 50 1 − t + 1 1 − t = − ( 1 + t + ⋯ + t 49 ) + 1 1 − t \frac{t^{50}}{1-t} = -\frac{1-t^{50}}{1-t} + \frac{1}{1-t} = -(1+t+\dots+t^{49}) + \frac{1}{1-t} 1 − t t 50 = − 1 − t 1 − t 50 + 1 − t 1 = − ( 1 + t + ⋯ + t 49 ) + 1 − t 1 ∫ 0 α 1 1 − t d t = − ln ( 1 − α ) = − β \int_0^\alpha \frac{1}{1-t}\,dt = -\ln(1-\alpha) = -\beta ∫ 0 α 1 − t 1 d t = − ln ( 1 − α ) = − β
B · P 50 ( α ) − β P_{50}(\alpha)-\beta P 50 ( α ) − β Applying the negative sign to β \beta β ∫ 0 α 1 1 − t d t = − β \int_0^\alpha \frac{1}{1-t}\,dt = -\beta ∫ 0 α 1 − t 1 d t = − β P 50 ( α ) P_{50}(\alpha) P 50 ( α ) t 50 − 1 1 − t = − ( 1 + t + ⋯ + t 49 ) \frac{t^{50}-1}{1-t} = -(1+t+\dots+t^{49}) 1 − t t 50 − 1 = − ( 1 + t + ⋯ + t 49 ) − P 50 ( α ) -P_{50}(\alpha) − P 50 ( α ) + P 50 ( α ) +P_{50}(\alpha) + P 50 ( α )
C · β − P 50 ( α ) \beta-P_{50}(\alpha) β − P 50 ( α ) Incorrectly evaluating ∫ 0 α 1 1 − t d t \int_0^\alpha \frac{1}{1-t}\,dt ∫ 0 α 1 − t 1 d t + ln ( 1 − α ) = β +\ln(1-\alpha) = \beta + ln ( 1 − α ) = β − ln ( 1 − α ) = − β -\ln(1-\alpha) = -\beta − ln ( 1 − α ) = − β ∫ 0 α 1 1 − t d t \int_0^\alpha \frac{1}{1-t}\,dt ∫ 0 α 1 − t 1 d t [ − ln ( 1 − t ) ] 0 α = − ln ( 1 − α ) = − β [-\ln(1-t)]_0^\alpha = -\ln(1-\alpha) = -\beta [ − ln ( 1 − t ) ] 0 α = − ln ( 1 − α ) = − β ( 1 − t ) (1-t) ( 1 − t )
D · -(β \beta β α \alpha α None. This is the correct option. Correctly applying the division t 50 1 − t = − ( 1 + t + ⋯ + t 49 ) + 1 1 − t \frac{t^{50}}{1-t} = -(1+t+\dots+t^{49}) + \frac{1}{1-t} 1 − t t 50 = − ( 1 + t + ⋯ + t 49 ) + 1 − t 1 − P 50 ( α ) − β = − ( β + P 50 ( α ) ) -P_{50}(\alpha) - \beta = -(\beta + P_{50}(\alpha)) − P 50 ( α ) − β = − ( β + P 50 ( α ))
Step Working
01 given a l p h a i n ( 0 , 1 ) \\alpha \\in (0,1) a l p ha in ( 0 , 1 ) b e t a = l o g e ( 1 − a l p h a ) \\beta = \\log_e(1 - \\alpha) b e t a = l o g e ( 1 − a l p ha ) P n ( x ) = s u m k = 1 n f r a c x k k P_n(x) = \\sum_{k=1}^n \\frac{x^k}{k} P n ( x ) = s u m k = 1 n f r a c x k k x i n ( 0 , 1 ) x \\in (0,1) x in ( 0 , 1 )
02 goal Evaluate I = ∫ 0 α t 50 1 − t d t I = \int_0^\alpha \frac{t^{50}}{1 - t}\,dt I = ∫ 0 α 1 − t t 50 d t β \beta β P 50 ( α ) P_{50}(\alpha) P 50 ( α )
03 approach Rewrite the numerator t 50 t^{50} t 50 ( t 50 − 1 ) + 1 = − ( 1 − t 50 ) + 1 (t^{50} - 1) + 1 = -(1 - t^{50}) + 1 ( t 50 − 1 ) + 1 = − ( 1 − t 50 ) + 1 f r a c 1 − t 50 1 − t = 1 + t + t 2 + d o t s + t 49 \\frac{1 - t^{50}}{1 - t} = 1 + t + t^2 + \\dots + t^{49} f r a c 1 − t 50 1 − t = 1 + t + t 2 + d o t s + t 49
04 execute Split the integral into two parts: I = − ∫ 0 α 1 − t 50 1 − t d t + ∫ 0 α 1 1 − t d t = − ∫ 0 α ( 1 + t + t 2 + ⋯ + t 49 ) d t + [ − ln ( 1 − t ) ] 0 α I = -\int_0^\alpha \frac{1 - t^{50}}{1 - t}\,dt + \int_0^\alpha \frac{1}{1 - t}\,dt = -\int_0^\alpha \left(1 + t + t^2 + \dots + t^{49}\right)dt + \left[ -\ln(1 - t) \right]_0^\alpha I = − ∫ 0 α 1 − t 1 − t 50 d t + ∫ 0 α 1 − t 1 d t = − ∫ 0 α ( 1 + t + t 2 + ⋯ + t 49 ) d t + [ − ln ( 1 − t ) ] 0 α
05 execute Integrate term-by-term: − [ t + t 2 2 + ⋯ + t 50 50 ] 0 α − ln ( 1 − α ) = − P 50 ( α ) − β = − ( β + P 50 ( α ) ) -\left[ t + \frac{t^2}{2} + \dots + \frac{t^{50}}{50} \right]_0^\alpha - \ln(1 - \alpha) = -P_{50}(\alpha) - \beta = -(\beta + P_{50}(\alpha)) − [ t + 2 t 2 + ⋯ + 50 t 50 ] 0 α − ln ( 1 − α ) = − P 50 ( α ) − β = − ( β + P 50 ( α ))
✓ verify Check signs: for t i n ( 0 , a l p h a ) t \\in (0, \\alpha) t in ( 0 , a l p ha ) f r a c t 50 1 − t > 0 \\frac{t^{50}}{1-t} > 0 f r a c t 50 1 − t > 0 b e t a = l n ( 1 − a l p h a ) < 0 \\beta = \\ln(1 - \\alpha) < 0 b e t a = l n ( 1 − a l p ha ) < 0 P 50 ( a l p h a ) = s u m k = 1 50 f r a c a l p h a k k < − l n ( 1 − a l p h a ) = − b e t a P_{50}(\\alpha) = \\sum_{k=1}^{50} \\frac{\\alpha^k}{k} < -\\ln(1-\\alpha) = -\\beta P 50 ( a l p ha ) = s u m k = 1 50 f r a c a l p h a k k < − l n ( 1 − a l p ha ) = − b e t a b e t a + P 50 ( a l p h a ) < 0 \\beta + P_{50}(\\alpha) < 0 b e t a + P 50 ( a l p ha ) < 0 − ( b e t a + P 50 ( a l p h a ) ) > 0 -(\\beta + P_{50}(\\alpha)) > 0 − ( b e t a + P 50 ( a l p ha )) > 0
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Question type Single correct
Exam relevance JEE Main · Mathematics
Concepts assessed Mathematics
Academic status Reviewed by official_key
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Editorial review 8 September 2026 Quick checks
Students also ask Why not use integration by parts? Integration by parts on t^50 * (1-t)^(-1) would differentiate or integrate (1-t)^(-1) to produce higher powers or complex logarithmic integrals, creating a non-terminating chain. Algebraic division is direct and terminates immediately.
Answer Rewriting t 50 1 − t = t 50 − 1 1 − t + 1 1 − t = − ( 1 + t + ⋯ + t 49 ) + 1 1 − t \frac{t^{50}}{1-t} = \frac{t^{50}-1}{1-t} + \frac{1}{1-t} = -(1 + t + \dots + t^{49}) + \frac{1}{1-t} 1 − t t 50 = 1 − t t 50 − 1 + 1 − t 1 = − ( 1 + t + ⋯ + t 49 ) + 1 − t 1 − P 50 ( α ) − ln ( 1 − α ) = − ( β + P 50 ( α ) ) -P_{50}(\alpha) - \ln(1-\alpha) = -(\beta + P_{50}(\alpha)) − P 50 ( α ) − ln ( 1 − α ) = − ( β + P 50 ( α ))
Why each option works or fails A: β + P 50 ( α ) \beta+P_{50}(\alpha) β + P 50 ( α ) - Dropping the negative sign on both the polynomial terms and the logarithmic integral during integration. Observe that t 50 1 − t = − 1 − t 50 1 − t + 1 1 − t = − ( 1 + t + ⋯ + t 49 ) + 1 1 − t \frac{t^{50}}{1-t} = -\frac{1-t^{50}}{1-t} + \frac{1}{1-t} = -(1+t+\dots+t^{49}) + \frac{1}{1-t} 1 − t t 50 = − 1 − t 1 − t 50 + 1 − t 1 = − ( 1 + t + ⋯ + t 49 ) + 1 − t 1 ∫ 0 α 1 1 − t d t = − ln ( 1 − α ) = − β \int_0^\alpha \frac{1}{1-t}\,dt = -\ln(1-\alpha) = -\beta ∫ 0 α 1 − t 1 d t = − ln ( 1 − α ) = − β B: P 50 ( α ) − β P_{50}(\alpha)-\beta P 50 ( α ) − β - Applying the negative sign to β \beta β ∫ 0 α 1 1 − t d t = − β \int_0^\alpha \frac{1}{1-t}\,dt = -\beta ∫ 0 α 1 − t 1 d t = − β P 50 ( α ) P_{50}(\alpha) P 50 ( α ) t 50 − 1 1 − t = − ( 1 + t + ⋯ + t 49 ) \frac{t^{50}-1}{1-t} = -(1+t+\dots+t^{49}) 1 − t t 50 − 1 = − ( 1 + t + ⋯ + t 49 ) − P 50 ( α ) -P_{50}(\alpha) − P 50 ( α ) + P 50 ( α ) +P_{50}(\alpha) + P 50 ( α ) C: β − P 50 ( α ) \beta-P_{50}(\alpha) β − P 50 ( α ) - Incorrectly evaluating ∫ 0 α 1 1 − t d t \int_0^\alpha \frac{1}{1-t}\,dt ∫ 0 α 1 − t 1 d t + ln ( 1 − α ) = β +\ln(1-\alpha) = \beta + ln ( 1 − α ) = β − ln ( 1 − α ) = − β -\ln(1-\alpha) = -\beta − ln ( 1 − α ) = − β ∫ 0 α 1 1 − t d t \int_0^\alpha \frac{1}{1-t}\,dt ∫ 0 α 1 − t 1 d t [ − ln ( 1 − t ) ] 0 α = − ln ( 1 − α ) = − β [-\ln(1-t)]_0^\alpha = -\ln(1-\alpha) = -\beta [ − ln ( 1 − t ) ] 0 α = − ln ( 1 − α ) = − β ( 1 − t ) (1-t) ( 1 − t ) D · correct: -(β \beta β α \alpha α - None. This is the correct option. Correctly applying the division t 50 1 − t = − ( 1 + t + ⋯ + t 49 ) + 1 1 − t \frac{t^{50}}{1-t} = -(1+t+\dots+t^{49}) + \frac{1}{1-t} 1 − t t 50 = − ( 1 + t + ⋯ + t 49 ) + 1 − t 1 − P 50 ( α ) − β = − ( β + P 50 ( α ) ) -P_{50}(\alpha) - \beta = -(\beta + P_{50}(\alpha)) − P 50 ( α ) − β = − ( β + P 50 ( α )) Step-by-step solution given: a l p h a i n ( 0 , 1 ) \\alpha \\in (0,1) a l p ha in ( 0 , 1 ) b e t a = l o g e ( 1 − a l p h a ) \\beta = \\log_e(1 - \\alpha) b e t a = l o g e ( 1 − a l p ha ) P n ( x ) = s u m k = 1 n f r a c x k k P_n(x) = \\sum_{k=1}^n \\frac{x^k}{k} P n ( x ) = s u m k = 1 n f r a c x k k x i n ( 0 , 1 ) x \\in (0,1) x in ( 0 , 1 ) goal: Evaluate I = ∫ 0 α t 50 1 − t d t I = \int_0^\alpha \frac{t^{50}}{1 - t}\,dt I = ∫ 0 α 1 − t t 50 d t β \beta β P 50 ( α ) P_{50}(\alpha) P 50 ( α ) approach: Rewrite the numerator t 50 t^{50} t 50 ( t 50 − 1 ) + 1 = − ( 1 − t 50 ) + 1 (t^{50} - 1) + 1 = -(1 - t^{50}) + 1 ( t 50 − 1 ) + 1 = − ( 1 − t 50 ) + 1 f r a c 1 − t 50 1 − t = 1 + t + t 2 + d o t s + t 49 \\frac{1 - t^{50}}{1 - t} = 1 + t + t^2 + \\dots + t^{49} f r a c 1 − t 50 1 − t = 1 + t + t 2 + d o t s + t 49 execute: Split the integral into two parts: I = − ∫ 0 α 1 − t 50 1 − t d t + ∫ 0 α 1 1 − t d t = − ∫ 0 α ( 1 + t + t 2 + ⋯ + t 49 ) d t + [ − ln ( 1 − t ) ] 0 α I = -\int_0^\alpha \frac{1 - t^{50}}{1 - t}\,dt + \int_0^\alpha \frac{1}{1 - t}\,dt = -\int_0^\alpha \left(1 + t + t^2 + \dots + t^{49}\right)dt + \left[ -\ln(1 - t) \right]_0^\alpha I = − ∫ 0 α 1 − t 1 − t 50 d t + ∫ 0 α 1 − t 1 d t = − ∫ 0 α ( 1 + t + t 2 + ⋯ + t 49 ) d t + [ − ln ( 1 − t ) ] 0 α execute: Integrate term-by-term: − [ t + t 2 2 + ⋯ + t 50 50 ] 0 α − ln ( 1 − α ) = − P 50 ( α ) − β = − ( β + P 50 ( α ) ) -\left[ t + \frac{t^2}{2} + \dots + \frac{t^{50}}{50} \right]_0^\alpha - \ln(1 - \alpha) = -P_{50}(\alpha) - \beta = -(\beta + P_{50}(\alpha)) − [ t + 2 t 2 + ⋯ + 50 t 50 ] 0 α − ln ( 1 − α ) = − P 50 ( α ) − β = − ( β + P 50 ( α )) verify: Check signs: for t i n ( 0 , a l p h a ) t \\in (0, \\alpha) t in ( 0 , a l p ha ) f r a c t 50 1 − t > 0 \\frac{t^{50}}{1-t} > 0 f r a c t 50 1 − t > 0 b e t a = l n ( 1 − a l p h a ) < 0 \\beta = \\ln(1 - \\alpha) < 0 b e t a = l n ( 1 − a l p ha ) < 0 P 50 ( a l p h a ) = s u m k = 1 50 f r a c a l p h a k k < − l n ( 1 − a l p h a ) = − b e t a P_{50}(\\alpha) = \\sum_{k=1}^{50} \\frac{\\alpha^k}{k} < -\\ln(1-\\alpha) = -\\beta P 50 ( a l p ha ) = s u m k = 1 50 f r a c a l p h a k k < − l n ( 1 − a l p ha ) = − b e t a b e t a + P 50 ( a l p h a ) < 0 \\beta + P_{50}(\\alpha) < 0 b e t a + P 50 ( a l p ha ) < 0 − ( b e t a + P 50 ( a l p h a ) ) > 0 -(\\beta + P_{50}(\\alpha)) > 0 − ( b e t a + P 50 ( a l p ha )) > 0 Shortcut: When to use it: When familiar with the power series expansion of -ln(1-t).
given: For ∣ t ∣ < 1 |t| < 1 ∣ t ∣ < 1 -\\ln(1 - t) = \\sum_{k=1}^\\infty \\frac{t^k}{k} .
goal: Relate \\int_0^\\alpha \\frac{t^{50}}{1-t}dt to the truncation of the series.
execute: Note that \\frac{t^{50}}{1-t} = \\sum_{k=50}^\\infty t^k = t^{50} + t^{51} + \\dots . Integrating term by term from 0 0 0 a l p h a \\alpha a l p ha \\int_0^\\alpha \\frac{t^{50}}{1-t}dt = \\sum_{k=50}^\\infty \\frac{\\alpha^{k+1}}{k+1} = \\sum_{m=51}^\\infty \\frac{\\alpha^m}{m} .
execute: The full series is -\\beta = -\\ln(1-\\alpha) = \\sum_{m=1}^\\infty \\frac{\\alpha^m}{m} = P_{50}(\\alpha) + \\sum_{m=51}^\\infty \\frac{\\alpha^m}{m} . Therefore, the integral equals − b e t a − P 50 ( a l p h a ) = − ( b e t a + P 50 ( a l p h a ) ) -\\beta - P_{50}(\\alpha) = -(\\beta + P_{50}(\\alpha)) − b e t a − P 50 ( a l p ha ) = − ( b e t a + P 50 ( a l p ha ))
verify: Since P 50 ( a l p h a ) P_{50}(\\alpha) P 50 ( a l p ha ) − b e t a -\\beta − b e t a − b e t a − P 50 ( a l p h a ) -\\beta - P_{50}(\\alpha) − b e t a − P 50 ( a l p ha )