JEE MainMathematics

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Integral of $t^{n} / (1 - t)$ in Terms of Logarithmic Series

Let α ∈(0,1) and β=log_e(1-α). Let P_n(x)=x+(x^2)/2+(x^3)/3+…+(x^n)/n, x ∈(0,1). Then the integral integral from 0 to α of (t^50)/(1-t)dt is equal to

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Aβ+P_50(α)BP_50(α)-βCβ-P_50(α)D-(β +P_{50}(α))

Step-by-step solutionView

Correct answer

Rewriting $\frac{t^{50}}{1 - t} = \frac{t^{50} - 1}{1 - t} + \frac{1}{1 - t} = - (1 + t + \dots + t^{49}) + \frac{1}{1 - t}$ and integrating yields $- P_{50} (α) - \ln (1 - α) = - (β + P_{50} (α))$ .

Option analysis

Why each option works or fails

A · β+P_50(α)

Dropping the negative sign on both the polynomial terms and the logarithmic integral during integration. Observe that $\frac{t^{50}}{1 - t} = - \frac{1 - t^{50}}{1 - t} + \frac{1}{1 - t} = - (1 + t + \dots + t^{49}) + \frac{1}{1 - t}$ , and $\int_{0}^{α} \frac{1}{1 - t} d t = - \ln (1 - α) = - β$ .

B · P_50(α)-β

Applying the negative sign to $β$ from $\int_{0}^{α} \frac{1}{1 - t} d t = - β$ while incorrectly keeping $P_{50} (α)$ positive. Remember that $\frac{t^{50} - 1}{1 - t} = - (1 + t + \dots + t^{49})$ , which integrates to $- P_{50} (α)$ , not $+ P_{50} (α)$ .

C · β-P_50(α)

Incorrectly evaluating $\int_{0}^{α} \frac{1}{1 - t} d t$ as $+ \ln (1 - α) = β$ rather than $- \ln (1 - α) = - β$ . The integral $\int_{0}^{α} \frac{1}{1 - t} d t$ equals $[- \ln (1 - t)]_{0}^{α} = - \ln (1 - α) = - β$ due to the chain rule on $(1 - t)$ .

D · -(β +P_{50}(α))

This is the correct option. Correctly applying the division $\frac{t^{50}}{1 - t} = - (1 + t + \dots + t^{49}) + \frac{1}{1 - t}$ gives the integrated value $- P_{50} (α) - β = - (β + P_{50} (α))$ .

Reviewed route

Solution

StepWorking

01Given

$a l p h a i n (0, 1)$ , $b e t a = l o g_{e} (1 - a l p h a)$ , and $P_{n} (x) = s u m_{k = 1}^{n} f r a c x^{k} k$ for $x i n (0, 1)$ .

02Goal

Evaluate $I = \int_{0}^{α} \frac{t^{50}}{1 - t} d t$ in terms of $β$ and $P_{50} (α)$ .

03Approach

Rewrite the numerator $t^{50}$ as $(t^{50} - 1) + 1 = - (1 - t^{50}) + 1$ . Then use the finite geometric series identity $f r a c 1 - t^{50} 1 - t = 1 + t + t^{2} + d o t s + t^{49}$ to decompose the integrand.

04Execute

Split the integral into two parts: $I = - \int_{0}^{α} \frac{1 - t^{50}}{1 - t} d t + \int_{0}^{α} \frac{1}{1 - t} d t = - \int_{0}^{α} (1 + t + t^{2} + \dots + t^{49}) d t + {[- \ln (1 - t)]}_{0}^{α}$ .

05Execute

Integrate term-by-term: $- {[t + \frac{t^{2}}{2} + \dots + \frac{t^{50}}{50}]}_{0}^{α} - \ln (1 - α) = - P_{50} (α) - β = - (β + P_{50} (α))$ .

✓Verify

Check signs: for $t i n (0, a l p h a)$ , the integrand $f r a c t^{50} 1 - t > 0$ , so the integral must be positive. Since $b e t a = l n (1 - a l p h a) < 0$ and $P_{50} (a l p h a) = s u m_{k = 1}^{50} f r a c a l p h a^{k} k < - l n (1 - a l p h a) = - b e t a$ , we have $b e t a + P_{50} (a l p h a) < 0$ , making $- (b e t a + P_{50} (a l p h a)) > 0$ . This confirms the overall sign is strictly positive.

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
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 $\frac{t^{50}}{1 - t} = \frac{t^{50} - 1}{1 - t} + \frac{1}{1 - t} = - (1 + t + \dots + t^{49}) + \frac{1}{1 - t}$ and integrating yields $- P_{50} (α) - \ln (1 - α) = - (β + P_{50} (α))$ .

Why each option works or fails

A: β+P_50(α) - Dropping the negative sign on both the polynomial terms and the logarithmic integral during integration. Observe that $\frac{t^{50}}{1 - t} = - \frac{1 - t^{50}}{1 - t} + \frac{1}{1 - t} = - (1 + t + \dots + t^{49}) + \frac{1}{1 - t}$ , and $\int_{0}^{α} \frac{1}{1 - t} d t = - \ln (1 - α) = - β$ .
B: P_50(α)-β - Applying the negative sign to $β$ from $\int_{0}^{α} \frac{1}{1 - t} d t = - β$ while incorrectly keeping $P_{50} (α)$ positive. Remember that $\frac{t^{50} - 1}{1 - t} = - (1 + t + \dots + t^{49})$ , which integrates to $- P_{50} (α)$ , not $+ P_{50} (α)$ .
C: β-P_50(α) - Incorrectly evaluating $\int_{0}^{α} \frac{1}{1 - t} d t$ as $+ \ln (1 - α) = β$ rather than $- \ln (1 - α) = - β$ . The integral $\int_{0}^{α} \frac{1}{1 - t} d t$ equals $\int_{0}^{α} \frac{1}{1 - t} d t$ due to the chain rule on $(1 - t)$ .
D · correct: -(β +P_{50}(α)) - This is the correct option. Correctly applying the division $\frac{t^{50}}{1 - t} = - (1 + t + \dots + t^{49}) + \frac{1}{1 - t}$ gives the integrated value $- P_{50} (α) - β = - (β + P_{50} (α))$ .

Step-by-step solution

Given: $a l p h a i n (0, 1)$ , $b e t a = l o g_{e} (1 - a l p h a)$ , and $P_{n} (x) = s u m_{k = 1}^{n} f r a c x^{k} k$ for $x i n (0, 1)$ .
Goal: Evaluate $I = \int_{0}^{α} \frac{t^{50}}{1 - t} d t$ in terms of $β$ and $P_{50} (α)$ .
Approach: Rewrite the numerator $t^{50}$ as $(t^{50} - 1) + 1 = - (1 - t^{50}) + 1$ . Then use the finite geometric series identity $f r a c 1 - t^{50} 1 - t = 1 + t + t^{2} + d o t s + t^{49}$ to decompose the integrand.
Execute: Split the integral into two parts: $I = - \int_{0}^{α} \frac{1 - t^{50}}{1 - t} d t + \int_{0}^{α} \frac{1}{1 - t} d t = - \int_{0}^{α} (1 + t + t^{2} + \dots + t^{49}) d t + {[- \ln (1 - t)]}_{0}^{α}$ .
Execute: Integrate term-by-term: $- {[t + \frac{t^{2}}{2} + \dots + \frac{t^{50}}{50}]}_{0}^{α} - \ln (1 - α) = - P_{50} (α) - β = - (β + P_{50} (α))$ .
Verify: Check signs: for $t i n (0, a l p h a)$ , the integrand $f r a c t^{50} 1 - t > 0$ , so the integral must be positive. Since $b e t a = l n (1 - a l p h a) < 0$ and $P_{50} (a l p h a) = s u m_{k = 1}^{50} f r a c a l p h a^{k} k < - l n (1 - a l p h a) = - b e t a$ , we have $b e t a + P_{50} (a l p h a) < 0$ , making $- (b e t a + P_{50} (a l p h a)) > 0$ . This confirms the overall sign is strictly positive.

Shortcut:

When to use it: When familiar with the power series expansion of -ln(1-t).

Given: For $∣ t ∣ < 1$ , the infinite series is $-\\ln(1 - t) = \\sum_{k=1}^\\infty \\frac{t^k}{k}$ .

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 $0$ to $a l p h a$ : $\\int_0^\\alpha \\frac{t^{50}}{1-t}dt = \\sum_{k=50}^\\infty \\frac{\\alpha^{k+1}}{k+1} = \\sum_{m=51}^\\infty \\frac{\\alpha^m}{m}$ .

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 $b e t a + P_{50} (a l p h a) < 0$ .

Verify: Since $P_{50} (a l p h a)$ is the 50th partial sum of $- b e t a$ , the tail must equal $- b e t a - P_{50} (a l p h a)$ , matching Option 3 exactly.

Integral of $t^{n} / (1 - t)$ in Terms of Logarithmic Series

What feels right?

Rewriting $\frac{t^{50}}{1 - t} = \frac{t^{50} - 1}{1 - t} + \frac{1}{1 - t} = - (1 + t + \dots + t^{49}) + \frac{1}{1 - t}$ and integrating yields $- P_{50} (α) - \ln (1 - α) = - (β + P_{50} (α))$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

✓Verify

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