If $$\int \frac{\sin x}{\sin^3 x + \cos^3 x} dx = \alpha\log_e |1 + \tan x| + \beta\log_e|1 - \tan x + \tan^2 x| + \gamma\tan^{-1}\frac{2\tan x - 1}{\sqrt{3}} + C$$, when $$C$$ is constant of integration, then the value of $$18\left(\alpha+\beta+\gamma^2\right)$$ is _________.

Divide both the numerator and the denominator by $$\cos^3 x$$:

$$I = \int \frac{\frac{\sin x}{\cos^3 x}}{\frac{\sin^3 x}{\cos^3 x} + \frac{\cos^3 x}{\cos^3 x}} dx = \int \frac{\tan x \cdot \sec^2 x}{\tan^3 x + 1} dx$$

Let $$u = \tan x$$. Then $$du = \sec^2 x \, dx$$. The integral becomes:

$$I = \int \frac{u}{u^3 + 1} du$$

The denominator factors as $$u^3 + 1 = (u+1)(u^2 - u + 1)$$. We set up the partial fractions:

$$\frac{u}{(u+1)(u^2 - u + 1)} = \frac{A}{u+1} + \frac{Bu + D}{u^2 - u + 1}$$

Multiplying through by the denominator:

$$u = A(u^2 - u + 1) + (Bu + D)(u + 1)$$

Comparing coefficients:

• $$u^2$$ terms: $$A + B = 0 \implies B = -A$$
• Constant terms: $$A + D = 0 \implies D = -A$$
• $$u$$ terms: $$-A + B + D = 1 \implies -A - A - A = 1 \implies A = -1/3$$

Thus:

$$A = -\frac{1}{3}, \quad B = \frac{1}{3}, \quad D = \frac{1}{3}$$

$$\frac{u}{u^3 + 1} = -\frac{1}{3(u+1)} + \frac{1}{3} \left( \frac{u+1}{u^2 - u + 1} \right)$$

Integrating term by term:

$$I = -\frac{1}{3} \ln|u+1| + \frac{1}{3} \int \frac{u+1}{u^2 - u + 1} du$$

For the second integral, we create the derivative of the denominator ($$2u-1$$) in the numerator:

$$\frac{u+1}{u^2-u+1} = \frac{1}{2} \left( \frac{2u-1+3}{u^2-u+1} \right) = \frac{1}{2} \frac{2u-1}{u^2-u+1} + \frac{3}{2} \frac{1}{(u-1/2)^2 + (\sqrt{3}/2)^2}$$

$$I = -\frac{1}{3} \ln|u+1| + \frac{1}{6} \ln|u^2-u+1| + \frac{1}{2} \cdot \frac{1}{\sqrt{3}/2} \tan^{-1} \left( \frac{u-1/2}{\sqrt{3}/2} \right)$$

$$I = -\frac{1}{3} \ln|1+\tan x| + \frac{1}{6} \ln|1-\tan x + \tan^2 x| + \frac{1}{\sqrt{3}} \tan^{-1} \left( \frac{2\tan x - 1}{\sqrt{3}} \right) + C$$

By comparing with the expression in the question:

• $$\alpha = -\frac{1}{3}$$
• $$\beta = \frac{1}{6}$$
• $$\gamma = \frac{1}{\sqrt{3}}$$

$$18\alpha + 18\beta + 18\gamma^2 = 18\left(-\frac{1}{3}\right) + 18\left(\frac{1}{6}\right) + 18\left(\frac{1}{3}\right) = -6 + 3 + 6 = 3$$

The correct answer is 3