1. **State the problem:** We need to evaluate the integral $$\int \left( \frac{3}{\sqrt{1 - x^2}} + \frac{4}{\sqrt{1 + x^2}} + \frac{5x}{\sqrt{x^2 - 1}} \right) dx$$ 2. **Recall formulas and rules:** - The integral of $\frac{1}{\sqrt{1 - x^2}}$ is $\arcsin x + C$. - The integral of $\frac{1}{\sqrt{1 + x^2}}$ is $\sinh^{-1} x + C$ or $\ln|x + \sqrt{1+x^2}| + C$. - For the term $\frac{x}{\sqrt{x^2 - 1}}$, use substitution or recognize it as derivative of $\sqrt{x^2 - 1}$. 3. **Break the integral into three parts:** $$I = \int \frac{3}{\sqrt{1 - x^2}} dx + \int \frac{4}{\sqrt{1 + x^2}} dx + \int \frac{5x}{\sqrt{x^2 - 1}} dx$$ 4. **Evaluate each integral separately:** - First integral: $$\int \frac{3}{\sqrt{1 - x^2}} dx = 3 \int \frac{1}{\sqrt{1 - x^2}} dx = 3 \arcsin x + C_1$$ - Second integral: $$\int \frac{4}{\sqrt{1 + x^2}} dx = 4 \int \frac{1}{\sqrt{1 + x^2}} dx = 4 \sinh^{-1} x + C_2 = 4 \ln|x + \sqrt{1 + x^2}| + C_2$$ - Third integral: Let $u = x^2 - 1$, then $du = 2x dx$ so $x dx = \frac{du}{2}$. Rewrite the integral: $$\int \frac{5x}{\sqrt{x^2 - 1}} dx = 5 \int \frac{x}{\sqrt{u}} dx = 5 \int \frac{1}{\sqrt{u}} \cdot x dx = 5 \int \frac{1}{\sqrt{u}} \cdot \frac{du}{2} = \frac{5}{2} \int u^{-1/2} du$$ Integrate: $$\frac{5}{2} \cdot 2 u^{1/2} + C_3 = 5 \sqrt{u} + C_3 = 5 \sqrt{x^2 - 1} + C_3$$ 5. **Combine all results:** $$\int \left( \frac{3}{\sqrt{1 - x^2}} + \frac{4}{\sqrt{1 + x^2}} + \frac{5x}{\sqrt{x^2 - 1}} \right) dx = 3 \arcsin x + 4 \ln|x + \sqrt{1 + x^2}| + 5 \sqrt{x^2 - 1} + C$$ where $C = C_1 + C_2 + C_3$ is the constant of integration. This completes the solution.