The value of b > 3 for which 12\int\limits_3^b 1 \over (x^2 - 1)(x^2 - 4)dx = \log _e( 49 \over 40 ) , is equal to \underline\hspace2cm .

Definite Integration PYQ — JEE Main 2022 · 25 Jun · Shift 2 · Q39 with Solution

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Correct answer: 6

We need to solve $12\int_3^b \frac{1}{(x^2-1)(x^2-4)}\,dx=\ln\left(\frac{49}{40}\right), \qquad b>3.$
First simplify the integrand using partial fractions.

Since $(x^2-1)(x^2-4)=(x-1)(x+1)(x-2)(x+2),$ we try a decomposition in terms of $x^2$ : $\frac{1}{(x^2-1)(x^2-4)}=\frac{A}{x^2-1}+\frac{B}{x^2-4}.$

Then $1=A(x^2-4)+B(x^2-1).$ So, $A+B=0, \qquad -4A-B=1.$ From $B=-A$ , we get

\qquad B=\frac13.$$ Hence $$\frac{1}{(x^2-1)(x^2-4)}=\frac13\left(\frac{1}{x^2-4}-\frac{1}{x^2-1}\right).$$ Therefore, $$12\int_3^b \frac{1}{(x^2-1)(x^2-4)}dx =4\int_3^b \left(\frac{1}{x^2-4}-\frac{1}{x^2-1}\right)dx.$$ 3. Now integrate each term. Using $$\int \frac{dx}{x^2-a^2}=\frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right|+C,$$ we get $$\int \frac{dx}{x^2-4}=\frac14\ln\left|\frac{x-2}{x+2}\right|+C,$$ $$\int \frac{dx}{x^2-1}=\frac12\ln\left|\frac{x-1}{x+1}\right|+C.$$ So, $$4\int \left(\frac{1}{x^2-4}-\frac{1}{x^2-1}\right)dx =4\left[\frac14\ln\left|\frac{x-2}{x+2}\right| - \frac12\ln\left|\frac{x-1}{x+1}\right|\right].$$ Thus an antiderivative is $$\ln\left|\frac{x-2}{x+2}\right|-2\ln\left|\frac{x-1}{x+1}\right|.$$ Since $x>3$ in the interval of integration, absolute values may be dropped: $$F(x)=\ln\left(\frac{x-2}{x+2}\right)-2\ln\left(\frac{x-1}{x+1}\right).$$ 4. Evaluate the definite integral: $$12\int_3^b \frac{1}{(x^2-1)(x^2-4)}dx=F(b)-F(3).$$ Now, $$F(3)=\ln\left(\frac{1}{5}\right)-2\ln\left(\frac{2}{4}\right) =\ln\left(\frac{1}{5}\right)-2\ln\left(\frac12\right).$$ Since $2\ln(1/2)=\ln(1/4)$, $$F(3)=\ln\left(\frac{1}{5}\right)-\ln\left(\frac14\right) =\ln\left(\frac{4}{5}\right).$$ Hence $$F(b)-\ln\left(\frac45\right)=\ln\left(\frac{49}{40}\right).$$ So $$F(b)=\ln\left(\frac{49}{40}\cdot \frac45\right)=\ln\left(\frac{49}{50}\right).$$ Therefore, $$\ln\left(\frac{b-2}{b+2}\right)-2\ln\left(\frac{b-1}{b+1}\right)=\ln\left(\frac{49}{50}\right).$$ Combining logs, $$\ln\left[\frac{b-2}{b+2}\left(\frac{b+1}{b-1}\right)^2\right]=\ln\left(\frac{49}{50}\right).$$ Thus, $$\frac{b-2}{b+2}\left(\frac{b+1}{b-1}\right)^2=\frac{49}{50}.$$ 5. Now test the stored answer $b=6$: $$\frac{6-2}{6+2}\left(\frac{6+1}{6-1}\right)^2 =\frac48\left(\frac75\right)^2 =\frac12\cdot\frac{49}{25} =\frac{49}{50},$$ which satisfies the equation. Since the integral function is continuous and strictly increasing in $b$ for $b>3$ (integrand is positive), the solution is unique. 6. Therefore, $$\boxed{6}$$

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View written solutionFree

Correct answer: 6

We need to solve $12\int_3^b \frac{1}{(x^2-1)(x^2-4)}\,dx=\ln\left(\frac{49}{40}\right), \qquad b>3.$
First simplify the integrand using partial fractions.

Since $(x^2-1)(x^2-4)=(x-1)(x+1)(x-2)(x+2),$ we try a decomposition in terms of $x^2$ : $\frac{1}{(x^2-1)(x^2-4)}=\frac{A}{x^2-1}+\frac{B}{x^2-4}.$

Then $1=A(x^2-4)+B(x^2-1).$ So, $A+B=0, \qquad -4A-B=1.$ From $B=-A$ , we get

\qquad B=\frac13.$$ Hence $$\frac{1}{(x^2-1)(x^2-4)}=\frac13\left(\frac{1}{x^2-4}-\frac{1}{x^2-1}\right).$$ Therefore, $$12\int_3^b \frac{1}{(x^2-1)(x^2-4)}dx =4\int_3^b \left(\frac{1}{x^2-4}-\frac{1}{x^2-1}\right)dx.$$ 3. Now integrate each term. Using $$\int \frac{dx}{x^2-a^2}=\frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right|+C,$$ we get $$\int \frac{dx}{x^2-4}=\frac14\ln\left|\frac{x-2}{x+2}\right|+C,$$ $$\int \frac{dx}{x^2-1}=\frac12\ln\left|\frac{x-1}{x+1}\right|+C.$$ So, $$4\int \left(\frac{1}{x^2-4}-\frac{1}{x^2-1}\right)dx =4\left[\frac14\ln\left|\frac{x-2}{x+2}\right| - \frac12\ln\left|\frac{x-1}{x+1}\right|\right].$$ Thus an antiderivative is $$\ln\left|\frac{x-2}{x+2}\right|-2\ln\left|\frac{x-1}{x+1}\right|.$$ Since $x>3$ in the interval of integration, absolute values may be dropped: $$F(x)=\ln\left(\frac{x-2}{x+2}\right)-2\ln\left(\frac{x-1}{x+1}\right).$$ 4. Evaluate the definite integral: $$12\int_3^b \frac{1}{(x^2-1)(x^2-4)}dx=F(b)-F(3).$$ Now, $$F(3)=\ln\left(\frac{1}{5}\right)-2\ln\left(\frac{2}{4}\right) =\ln\left(\frac{1}{5}\right)-2\ln\left(\frac12\right).$$ Since $2\ln(1/2)=\ln(1/4)$, $$F(3)=\ln\left(\frac{1}{5}\right)-\ln\left(\frac14\right) =\ln\left(\frac{4}{5}\right).$$ Hence $$F(b)-\ln\left(\frac45\right)=\ln\left(\frac{49}{40}\right).$$ So $$F(b)=\ln\left(\frac{49}{40}\cdot \frac45\right)=\ln\left(\frac{49}{50}\right).$$ Therefore, $$\ln\left(\frac{b-2}{b+2}\right)-2\ln\left(\frac{b-1}{b+1}\right)=\ln\left(\frac{49}{50}\right).$$ Combining logs, $$\ln\left[\frac{b-2}{b+2}\left(\frac{b+1}{b-1}\right)^2\right]=\ln\left(\frac{49}{50}\right).$$ Thus, $$\frac{b-2}{b+2}\left(\frac{b+1}{b-1}\right)^2=\frac{49}{50}.$$ 5. Now test the stored answer $b=6$: $$\frac{6-2}{6+2}\left(\frac{6+1}{6-1}\right)^2 =\frac48\left(\frac75\right)^2 =\frac12\cdot\frac{49}{25} =\frac{49}{50},$$ which satisfies the equation. Since the integral function is continuous and strictly increasing in $b$ for $b>3$ (integrand is positive), the solution is unique. 6. Therefore, $$\boxed{6}$$

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