If ∫_0^1 4 cot^-1 (1-2 x+4 x^2 ) d x=atan^-1(2)-blog_e(5) , where a, b ∈ N , then (2 a+b) is equal to \_\_\_\_

Question

If ∫_0^1 4 cot^-1 (1-2 x+4 x^2 ) d x=atan^-1(2)-blog_e(5) , where a, b ∈ N , then (2 a+b) is equal to \_\_\_\_ .

MARKS Content Team · Accepted Answer

Correct answer: 9. Let I=∫_0^1 cot^-1 (1-2 x+4 x^2 ) d x I=∫_0^1 (cot^-1(2 x-1)-cot^-1(2 x) ) d x Applying king I=∫_0^1 (-cot^-1(2 x-1)+cot^-1(2 x-2) ) d x From (1) and (2) 2 I=∫_0^1 (cot^-1(2 x-2)-cot^-1(2 x) ) d x =∫_0^1 cot^-1(2 x-2) d x-∫_0^1 cot^-1(2 x) d x Applying King =∫_0^1 cot^-1(-2 x) d x-∫_0^1 cot^-1(2 x) d x =∫_0^1 (π-cot^-1(2 x) ) d x-∫_0^1 cot^-1(2 x) d x =∫_0^1 (π-2 cot^-1(2 x) ) d x =π-2 ∫_0^1 (cot^-1 2 x ) · 1 d x By parts =π-2 [ (x cot^-1 2 x )_0^1+∫_0^1 (2 x)/(1+4 x^2) d x ] Let 1+4 x^2=t 8 xdx=dt =π-2 [cot^-1 2+1/4 ∫_1^5 dt/t ] =π-2 cot^-1 2-1/2 ℓ n 5 2 I=2 tan^-1 2-1/2 ℓ n 5 ⇒ 4 I=4 tan^-1 2-ℓ n 5 ∴ 2 a+b=8+1=9

If $\int_{0}^{1} 4 \cot ^{-1}\left(1-2 x+4 x^{2}\right) \mathrm{d} x=\mathrm{atan}^{-1}(2)-\mathrm{blog}_{\mathrm{e}}(5)$, where $\mathrm{a}, \mathrm{b} \in \mathbf{N}$, then $(2 \mathrm{a}+\mathrm{b})$ is equal to $\_\_\_\_$.

Solution

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