The value of \(b>3\) for which \(12 \int_{3}^{b} \frac{1}{(x^{2}-1)(x^{2}-4)} dx=\log_{e}\left(\frac{49}{40}\right)\), is equal to

Definite Integrals — JEE Main · 2022 · 25 Jun 2022 · Shift 2 Question with Solution

The value of

b > 3

for which

12 \int_{3}^{b} \frac{1}{( x ^{2} - 1 ) ( x ^{2} - 4 )} d x = lo g_{e} (\frac{49}{40})

, is equal to

Answer: 6

Use partial fractions:

\frac{1}{( x ^{2} - 1 ) ( x ^{2} - 4 )} = \frac{1}{3} (\frac{1}{x ^{2} - 4} - \frac{1}{x ^{2} - 1})

. So

12 \int_{3}^{b} \frac{1}{( x ^{2} - 1 ) ( x ^{2} - 4 )} d x = 4 \int_{3}^{b} (\frac{1}{x ^{2} - 4} - \frac{1}{x ^{2} - 1}) d x

. Now

\int \frac{1}{x ^{2} - 4} d x = \frac{1}{4} ln \frac{x - 2}{x + 2}

and

\int \frac{1}{x ^{2} - 1} d x = \frac{1}{2} ln \frac{x - 1}{x + 1}

. So the integral becomes

4 [\frac{1}{4} ln \frac{x - 2}{x + 2} - \frac{1}{2} ln \frac{x - 1}{x + 1}]_{3}^{b} = [ln \frac{x - 2}{x + 2} - 2 ln \frac{x - 1}{x + 1}]_{3}^{b} = ln \frac{( b - 2 ) ( b + 1 ) ^{2}}{( b + 2 ) ( b - 1 ) ^{2}} - ln \frac{1 \cdot 16}{5 \cdot 4} = ln \frac{( b - 2 ) ( b + 1 ) ^{2}}{( b + 2 ) ( b - 1 ) ^{2}} - ln \frac{4}{5}

. Setting this equal to

ln \frac{49}{40} = ln \frac{49}{40}

, we get

ln \frac{( b - 2 ) ( b + 1 ) ^{2}}{( b + 2 ) ( b - 1 ) ^{2}} = ln \frac{49}{40} + ln \frac{4}{5} = ln \frac{49}{50}

. So

\frac{( b - 2 ) ( b + 1 ) ^{2}}{( b + 2 ) ( b - 1 ) ^{2}} = \frac{49}{50}

. Testing

b = 6

\frac{4 \cdot 49}{8 \cdot 25} = \frac{196}{200} = \frac{49}{50}

. Thus

b = 6

JEE Main · 2022 · 25 Jun 2022 · Shift 2 · Integer answer