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Definite Integration: Let Even Function Value Represents Greatest Integer Function

JEE Maths question with a full step-by-step solution.

Question
Let f(x)f(x) be an even function such that f(x)+f(x−2)=x2+(1−2x)f(x)+f\left(x-2\right) = x^2+\left(1-2x\right) ∀ x∈R\forall\,x \in \mathbb R and
∫02(f(x)x2−2x−1)dx=1+aln⁡(b),\int_0^2\left(\frac{f(x)}{x^2-2x-1}\right)dx = 1+a\ln\left(b\right),
 then the value of [a2]+[b]\left[a^2\right]+\left[b\right] is (where [ . ][\,.\,] represents the greatest integer function)
Solution
Answer: 2
Step 1: The right-hand side is x2−2x+1=(x−1)2x^2-2x+1 = \left(x-1\right)^2. Taking f(x)=x2−12f(x) = \tfrac{x^2-1}2, which is even,
f(x)+f(x−2)=x2−12+(x−2)2−12=x2−1+x2−4x+4−12=2x2−4x+22f(x)+f\left(x-2\right) = \frac{x^2-1}2+\frac{\left(x-2\right)^2-1}2 = \frac{x^2-1+x^2-4x+4-1}2 = \frac{2x^2-4x+2}2
 =(x−1)2= \left(x-1\right)^2 Step 2: x2−2x−1x^2-2x-1 has roots 1±2=−0.4141\pm\sqrt2 = -0.414 and 2.4142.414, both outside [0,2]\left[0,2\right], so the integrand is continuous on the whole range:
I=∫02(x2−1)/2x2−2x−1 dx=12∫02x2−1x2−2x−1 dxI = \int_0^2\frac{\left(x^2-1\right)/2}{x^2-2x-1}\,dx = \frac12\int_0^2\frac{x^2-1}{x^2-2x-1}\,dx
 Step 3: Splitting the numerator,
x2−1=(x2−2x−1)+2x=(x2−2x−1)+(2x−2)+2x^2-1 = \left(x^2-2x-1\right)+2x = \left(x^2-2x-1\right)+\left(2x-2\right)+2
I=12∫02[1+2x−2x2−2x−1+2(x−1)2−2]dxI = \frac12\int_0^2\left[1+\frac{2x-2}{x^2-2x-1}+\frac2{\left(x-1\right)^2-2}\right]dx
 Step 4:
∫021 dx=2;\int_0^2 1\,dx = 2 ;
∫022x−2x2−2x−1dx=[ln⁡∣x2−2x−1∣]02=ln⁡1−ln⁡1=0;\int_0^2\frac{2x-2}{x^2-2x-1}dx = \Big[\ln\left|x^2-2x-1\right|\Big]_0^2 = \ln1-\ln1 = 0 ;
 for the third, put u=x−1∈[−1,1]u = x-1 \in \left[-1,1\right] and use ∫duu2−2=122ln⁡∣u−2u+2∣\int\frac{du}{u^2-2} = \frac1{2\sqrt2}\ln\left|\frac{u-\sqrt2}{u+\sqrt2}\right|; the integrand 1u2−2\frac1{u^2-2} is even, so
∫−11duu2−2=2∫01duu2−2=222ln⁡2−12+1=12ln⁡(2−1)2=2 ln⁡(2−1)\int_{-1}^{1}\frac{du}{u^2-2} = 2\int_0^1\frac{du}{u^2-2} = \frac2{2\sqrt2}\ln\frac{\sqrt2-1}{\sqrt2+1} = \frac1{\sqrt2}\ln\left(\sqrt2-1\right)^2 = \sqrt2\,\ln\left(\sqrt2-1\right)
 So the third piece contributes 22ln⁡(2−1)2\sqrt2\ln\left(\sqrt2-1\right). Step 5:
I=12[2+0+22ln⁡(2−1)]=1+2 ln⁡(2−1)I = \frac12\left[2+0+2\sqrt2\ln\left(\sqrt2-1\right)\right] = 1+\sqrt2\,\ln\left(\sqrt2-1\right)
a=2,b=2−1=0.41421…a = \sqrt2 ,\qquad b = \sqrt2-1 = 0.41421\ldots
[a2]+[b]=[2]+[0.41421]=2+0=2\left[a^2\right]+\left[b\right] = \left[2\right]+\left[0.41421\right] = 2+0 = 2
 Answer: 22
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