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Mathematics Question on types of differential equations

The value of b > 3 for which 12∫3b1(x2−1)(x2−4)dx=loge(4040)12∫_3^b \frac{1}{(x^2-1)(x^2-4)}dx=log_e(\frac{40}{40}) is equal to.

Answer

The correct answer is: 6

l=∫3b1(x2−1)(x2−4)dx=13(1x2−4−1x2−1)dxl=∫_3^b \frac{1}{(x^2-1)(x^2-4)}dx=\frac{1}{3}(\frac{1}{x^2-4}-\frac{1}{x^2-1})dx

=ln((b−2b+2)(b+1)2b−12)−(In 45)=ln((\frac{b-2}{b+2})\frac{(b+1)^2}{b-1}^2)-(In\,\frac{4}{5})

After simplification ,

4940=(b−2)(b+2)(b+1)2(b−1)2.54\frac{49}{40}=\frac{(b-2)}{(b+2)}\frac{(b+1)^2}{(b-1)^2}.\frac{5}{4}

⇒ b = 6