$$\text{If } \int \frac{x + (\cos^{-1}3x)^2}{\sqrt{1-9x^2}} dx = P\sqrt{1-9x^2} + Q(\cos^{-1}3x)^3 + c \text{ then:}$$
Step-by-Step Solution
Key Concept: Use substitution strategically on each term separately: $u$-substitution for the rational part and inverse trigonometric substitution for the squared inverse function.
Split the integral into two parts: $\int \frac{x}{\sqrt{1-9x^2}} dx + \int \frac{(\cos^{-1}3x)^2}{\sqrt{1-9x^2}} dx$. For the first part, substitute $u = 1-9x^2$, so $du = -18x\,dx$, giving $-\frac{1}{18}\int u^{-1/2}du = -\frac{1}{9}\sqrt{1-9x^2}$. For the second part, substitute $t = \cos^{-1}3x$, so $dt = \frac{-3}{\sqrt{1-9x^2}}dx$, giving $-\frac{1}{3}\int t^2 dt = -\frac{1}{9}t^3 = -\frac{1}{9}(\cos^{-1}3x)^3$. Therefore $P = -\frac{1}{9}$ and $Q = -\frac{1}{9}$, making options 1 and 4 correct.
Correct Answer: 1,4