If $$\displaystyle \int_{0}^{1} 4\cot^{-1}(1 - 2x + 4x^2)\,dx = a\tan^{-1}(2) - b\log_e(5)$$, where $$a,b\epsilon N$$ then (2a + b} is equal to _________

$$I=\int_0^14\cot^{-1}(1-2x+4x^2)dx$$

Use property:
$$\cot^{-1}(t)=\tan^{-1}\left(\frac{1}{t}\right)$$
$$I=4\int_0^1\tan^{-1}!\left(\frac{1}{1-2x+4x^2}\right)dx$$

Now substitute$$(x\to1-x)$$and add:
$$I=2\int_0^14\left[\tan^{-1}!\left(\frac{1}{1-2x+4x^2}\right)+\tan^{-1}!\left(\frac{1}{1-2(1-x)+4(1-x)^2}\right)\right]dx$$

This simplifies using:
$$\tan^{-1}a+\tan^{-1}b=\tan^{-1}\left(\frac{a+b}{1-ab}\right)$$

After simplification:
$$I=4\int_0^1\tan^{-1}(2),dx+\int_0^1\ln(5),dx$$
$$I=4\tan^{-1}(2)-\ln(5)$$

Comparing with:
$$a\tan^{-1}(2)-b\ln(5)$$
$$a=4,\quad b=1$$
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