1. **Problem:** Evaluate the integral $$\int \ln(3x - 2) \, dx$$ 2. **Formula and rules:** Use integration by parts formula: $$\int u \, dv = uv - \int v \, du$$ Choose: $$u = \ln(3x - 2), \quad dv = dx$$ Then: $$du = \frac{3}{3x - 2} dx, \quad v = x$$ 3. **Apply integration by parts:** $$\int \ln(3x - 2) \, dx = x \ln(3x - 2) - \int x \cdot \frac{3}{3x - 2} dx$$ 4. **Simplify the integral:** $$\int \frac{3x}{3x - 2} dx = \int \frac{3x - 2 + 2}{3x - 2} \cdot 3 dx = \int \left(3 + \frac{6}{3x - 2}\right) dx$$ 5. **Split and integrate:** $$\int 3 dx + \int \frac{6}{3x - 2} dx = 3x + 6 \int \frac{1}{3x - 2} dx$$ 6. **Integrate the last term:** $$6 \int \frac{1}{3x - 2} dx = 6 \cdot \frac{1}{3} \ln|3x - 2| = 2 \ln|3x - 2|$$ 7. **Combine all parts:** $$\int \ln(3x - 2) dx = x \ln(3x - 2) - (3x + 2 \ln|3x - 2|) + C = x \ln(3x - 2) - 3x - 2 \ln|3x - 2| + C$$ --- 1. **Problem:** Evaluate the integral $$\int \ln(x^2 + 4) \, dx$$ 2. **Formula and rules:** Use integration by parts: $$u = \ln(x^2 + 4), \quad dv = dx$$ Then: $$du = \frac{2x}{x^2 + 4} dx, \quad v = x$$ 3. **Apply integration by parts:** $$\int \ln(x^2 + 4) dx = x \ln(x^2 + 4) - \int x \cdot \frac{2x}{x^2 + 4} dx = x \ln(x^2 + 4) - 2 \int \frac{x^2}{x^2 + 4} dx$$ 4. **Simplify the integral:** Rewrite the integrand: $$\frac{x^2}{x^2 + 4} = 1 - \frac{4}{x^2 + 4}$$ 5. **Split the integral:** $$\int \frac{x^2}{x^2 + 4} dx = \int 1 dx - 4 \int \frac{1}{x^2 + 4} dx = x - 4 \int \frac{1}{x^2 + 4} dx$$ 6. **Integrate the last term:** Recall: $$\int \frac{1}{x^2 + a^2} dx = \frac{1}{a} \arctan\left(\frac{x}{a}\right) + C$$ Here, $a=2$, so: $$\int \frac{1}{x^2 + 4} dx = \frac{1}{2} \arctan\left(\frac{x}{2}\right) + C$$ 7. **Combine all parts:** $$\int \ln(x^2 + 4) dx = x \ln(x^2 + 4) - 2 \left(x - 4 \cdot \frac{1}{2} \arctan\left(\frac{x}{2}\right)\right) + C = x \ln(x^2 + 4) - 2x + 4 \arctan\left(\frac{x}{2}\right) + C$$ --- 1. **Problem:** Evaluate the integral $$\int \sin^{-1} x \, dx$$ 2. **Formula and rules:** Use integration by parts: $$u = \sin^{-1} x, \quad dv = dx$$ Then: $$du = \frac{1}{\sqrt{1 - x^2}} dx, \quad v = x$$ 3. **Apply integration by parts:** $$\int \sin^{-1} x \, dx = x \sin^{-1} x - \int \frac{x}{\sqrt{1 - x^2}} dx$$ 4. **Simplify the integral:** Use substitution: Let $$t = 1 - x^2 \Rightarrow dt = -2x dx \Rightarrow -\frac{1}{2} dt = x dx$$ 5. **Rewrite the integral:** $$\int \frac{x}{\sqrt{1 - x^2}} dx = \int \frac{x}{\sqrt{t}} dx = -\frac{1}{2} \int t^{-1/2} dt = -\frac{1}{2} \cdot 2 t^{1/2} + C = -\sqrt{1 - x^2} + C$$ 6. **Combine all parts:** $$\int \sin^{-1} x \, dx = x \sin^{-1} x + \sqrt{1 - x^2} + C$$