If int_0^14cot^-1(1-2x+4x^2)dx=atan^-1(2)-blog_e(5)$\int_{0}^{1}4\cot^{-1}(1-2x+4x^{2})dx=a\tan^{-1}(2)-b\log_{e}(5)$, where a, b in N$a, b \in N$, then (2a+b)$(2a+b)$ is equal to ____.

### Related Formula cot^-1(y) = tan^-1left(frac1yright)$$\cot^{-1}(y) = \tan^{-1}\left(\frac{1}{y}\right)$$ tan^-1left(fracx - y1 + xyright) = tan^-1x - tan^-1y$$\tan^{-1}\left(\frac{x - y}{1 + xy}\right) = \tan^{-1}x - \tan^{-1}y$$ textKing's Property: int_0^a f(x)dx = int_0^a f(a-x)dx$$\text{King's Property: } \int_0^a f(x)dx = \int_0^a f(a-x)dx$$ ### Core Logic Let I = int_0^1 cot^-1(1-2x+4x^2) dx$I = \int_{0}^{1} \cot^{-1}(1-2x+4x^2) dx$. Convert cot^-1$\cot^{-1}$ to tan^-1$\tan^{-1}$: cot^-1(1 + 2x(2x-1)) = tan^-1left( frac11 + 2x(2x-1) right)$$\cot^{-1}(1 + 2x(2x-1)) = \tan^{-1}\left( \frac{1}{1 + 2x(2x-1)} \right)$$ Notice that 2x - (2x-1) = 1$2x - (2x-1) = 1$. Thus, the integrand is tan^-1left( frac2x - (2x-1)1 + 2x(2x-1) right)$\tan^{-1}\left( \frac{2x - (2x-1)}{1 + 2x(2x-1)} \right)$. I = int_0^1 left( tan^-1(2x) - tan^-1(2x-1) right) dx$$I = \int_0^1 \left( \tan^{-1}(2x) - \tan^{-1}(2x-1) \right) dx$$ ### Step 1: Apply Definite Integral Properties Applying King's property to the second term int_0^1 tan^-1(2x-1) dx$\int_0^1 \tan^{-1}(2x-1) dx$: x to 1-x implies int_0^1 tan^-1(2(1-x)-1) dx = int_0^1 tan^-1(1-2x) dx = -int_0^1 tan^-1(2x-1) dx$$x \to 1-x \implies \int_0^1 \tan^{-1}(2(1-x)-1) dx = \int_0^1 \tan^{-1}(1-2x) dx = -\int_0^1 \tan^{-1}(2x-1) dx$$ Wait, this implies int_0^1 tan^-1(2x-1) dx = 0$\int_0^1 \tan^{-1}(2x-1) dx = 0$! Thus, I = int_0^1 tan^-1(2x) dx$I = \int_0^1 \tan^{-1}(2x) dx$. ### Step 2: Integration by Parts Solve int_0^1 tan^-1(2x) cdot 1 \, dx$\int_0^1 \tan^{-1}(2x) \cdot 1 \, dx$: I = left[ x tan^-1(2x) right]_0^1 - int_0^1 x frac21+4x^2 dx$$I = \left[ x \tan^{-1}(2x) \right]_0^1 - \int_0^1 x \frac{2}{1+4x^2} dx$$ I = tan^-1(2) - frac14 int_0^1 frac8x1+4x^2 dx$$I = \tan^{-1}(2) - \frac{1}{4} \int_0^1 \frac{8x}{1+4x^2} dx$$ Let 1+4x^2 = t implies 8x dx = dt$1+4x^2 = t \implies 8x dx = dt$. At x=0, t=1$x=0, t=1$; at x=1, t=5$x=1, t=5$. I = tan^-1(2) - frac14 int_1^5 fracdtt = tan^-1(2) - frac14 ln(5)$$I = \tan^{-1}(2) - \frac{1}{4} \int_1^5 \frac{dt}{t} = \tan^{-1}(2) - \frac{1}{4} \ln(5)$$ ### Step 3: Compare and Calculate Result The original integral has a factor of 4$4$: 4I = 4tan^-1(2) - ln(5)$$4I = 4\tan^{-1}(2) - \ln(5)$$ Compare with atan^-1(2) - blog_e(5)$a\tan^{-1}(2) - b\log_e(5)$: a = 4, quad b = 1$$a = 4, \quad b = 1$$ Therefore, 2a + b = 2(4) + 1 = 9$2a + b = 2(4) + 1 = 9$. ### Pattern Recognition Always convert cot^-1$\cot^{-1}$ quadratic inputs into tan^-1fracx-y1+xy$\tan^{-1}\frac{x-y}{1+xy}$ forms. Apply King's property on symmetric limits; often one piece vanishes entirely. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Definite Integrals Class 12 Maths: Inverse Trigonometric Functions