If _log2^x1e^t-1 dt=π /6, then x=

B. 2log2. Step 1: Put e^t-1=u^2: e^t dt=2u du, so dt=2u duu^2+1 and the integrand becomes 2u^2+1. Step 2: Limits: t=log2 u=1; t=x u=e^x-1. The integral is 2tan^-1e^x-1-2⋅π /4. Step 3: 2tan^-1e^x-1=π /6+π /2=2π /3, so e^x-1=√(3), e^x=4 and x=2log2. Check: At x=2log2: 2tan^-1√(3)-2tan^-11=2π /3-π /2=π /6. Trap: Forgetting the lower-limit term 2tan^-11=π /2 gives tan^-1e^x-1=π /12, i.e. e^x 1.07, which matches no option.

Question 29 of 31 · Definite Integration

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NIMCET 2019

If $\int_{l o g 2}^{x} \frac{1}{e ^{t} - 1} d t = \frac{π}{6}$ , then $x =$

Answer (Detailed Solution Below)View answer ▾

Answer: Option B — $2 l o g 2$

Detailed Solution

Put $e^{t} - 1 = u^{2}$ : $e^{t} d t = 2 u d u$ , so $d t = \frac{2 u d u}{u ^{2} + 1}$ and the integrand becomes $\frac{2}{u ^{2} + 1}$ .

Limits: $t = l o g 2 \Rightarrow u = 1$ ; $t = x \Rightarrow u = e^{x} - 1$ . The integral is $2 t an^{- 1} e^{x} - 1 - 2 \cdot \frac{π}{4}$ .

$2 t an^{- 1} e^{x} - 1 = \frac{π}{6} + \frac{π}{2} = \frac{2 π}{3}$ , so $e^{x} - 1 = 3$ , $e^{x} = 4$ and $x = 2 l o g 2$ .

✓ Check

x = 2 l o g 2

2 t an^{- 1} 3 - 2 t an^{- 1} 1 = \frac{2 π}{3} - \frac{π}{2} = \frac{π}{6}

⚠ Common trap

Forgetting the lower-limit term

2 t an^{- 1} 1 = \frac{π}{2}

gives

t an^{- 1} e^{x} - 1 = \frac{π}{12}

, i.e.

e^{x} \approx 1.07

, which matches no option.

Quick tip

$\int \frac{d t}{e ^{t} - 1} = 2 tan^{- 1} e^{t} - 1$

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Definite Integration — Preparation

This question tests your understanding of Definite Integration, a key topic in the Quantitative Aptitude section. The detailed solution above uses a step-by-step approach to show the reasoning, and where available, a shortcut that helps you solve similar questions in under 90 seconds during the actual exam.

A definite integral gives a number rather than a family of functions. It is evaluated using the fundamental theorem of calculus: the integral of f(x) from a to b equals F(b) - F(a), where F is an antiderivative of f. This topic covers evaluating definite integrals, substitution with a change of limits, integration by parts with limits, and the definite integral as the limit of a sum. The properties of definite integrals are the key to most exam questions. The most useful is the "a + b - x" (or "a - x") property, which turns many difficult integrals into simple ones by adding the integral to itself. Also remember that the integral of an odd function over a symmetric interval from -a to a is zero, while the integral of an even function equals twice the integral from 0 to a. Practise the definite integration questions below, each with a step-by-step solution showing which property is used.

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