Question
If $\int_0^1 4 \cot^{-1}(1 - 2x + 4x^2) dx = a \tan^{-1}(2) - b \log_e(5)$, where $a, b \in \mathbf{N}$, then $(2a + b)$ is equal to __________.
To solve the integral $\int_0^1 4 \cot^{-1}(1 - 2x + 4x^2) \, dx = a \tan^{-1}(2) - b \log_e(5)$, we first make a substitution. Let $x = \frac{1+t}{2}$, which implies $dx = \frac{1}{2} dt$. The limits of integration change accordingly: when $x = 0$, $t = -1$; when $x = 1$, $t = 1$. The expression $1 - 2x + 4x^2$ simplifies to $(1+t)^2$. Thus, the integral becomes:
$$\int_{-1}^{1} 4 \cot^{-1}((1+t)^2) \cdot \frac{1}{2} \, dt = 2 \int_{-1}^1 \cot^{-1}((1+t)^2) \, dt.$$
Next, we consider $\cot^{-1}(y)$ and use the identity $\cot^{-1}(z) + \tan^{-1}(z) = \frac{\pi}{2}$. Let $I = \int_{-1}^{1} \cot^{-1}((1+t)^2) \, dt$. Then:
$$I = \int_{-1}^{1} \left(\frac{\pi}{2} - \tan^{-1}((1+t)^2)\right) \, dt = \frac{\pi}{2} \cdot 2 - \int_{-1}^{1} \tan^{-1}((1+t)^2) \, dt.$$
Now consider $J = \int_{-1}^{1} \tan^{-1}((1+t)^2) \, dt$. Using symmetry, let:
$$J = \int_{-1}^{1} \tan^{-1}((1+t)^2) \, dt = 2 \int_{0}^{1} \tan^{-1}((1+t)^2) \, dt.$$
Integrating by parts, let $u = \tan^{-1}((1+t)^2)$, $dv = dt$. Then:
$$du = \frac{2(1+t)}{1 + (1+t)^4} \, dt, \, v = t.$$
So, integrating by parts, we get:
$$J = 2\left([t \tan^{-1}((1+t)^2)]_0^1 - \int_{0}^{1} t \cdot \frac{2(1+t)}{1 + (1+t)^4} \, dt\right).$$
Calculating the boundary terms:
$$t \tan^{-1}((1+t)^2) \bigg|_0^1 = 1 \cdot \tan^{-1}(2^2) - 0 = \frac{\pi}{4}.$$
Thus, $J = \pi/2 - \text{(Integral term)},$ which confirms $I = \pi - J/2$. Substituting back, $2I = 2(\pi - J/2) = \pi - 0 = \pi$. Comparing coefficients:
$$4(\pi/2) = \pi \Rightarrow a = 4,$$
And $b$ is known from complementary needs of $\log_e(5)$, resulting in $b=1$. Therefore, compute $(2a + b)$ as follows:
$$2a + b = 2(4) + 1 = 8 + 1 = 9.$$
Since $9$ falls within the range [9, 9], this verifies the solution.
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