1. **State the problem:** We want to find the integral $$I = \int \ln(x^2 + 1) \, dx$$. 2. **Use integration by parts formula:** $$\int u \, dv = uv - \int v \, du$$. 3. **Choose parts:** Let $$u = \ln(x^2 + 1)$$ and $$dv = dx$$. 4. **Compute derivatives and integrals:** - $$du = \frac{d}{dx} \ln(x^2 + 1) = \frac{2x}{x^2 + 1} \, dx$$ - $$v = \int dx = x$$ 5. **Apply integration by parts:** $$I = x \ln(x^2 + 1) - \int x \cdot \frac{2x}{x^2 + 1} \, dx = x \ln(x^2 + 1) - 2 \int \frac{x^2}{x^2 + 1} \, dx$$ 6. **Simplify the integrand:** $$\frac{x^2}{x^2 + 1} = \frac{x^2 + 1 - 1}{x^2 + 1} = 1 - \frac{1}{x^2 + 1}$$ 7. **Rewrite the integral:** $$I = x \ln(x^2 + 1) - 2 \int \left(1 - \frac{1}{x^2 + 1}\right) dx = x \ln(x^2 + 1) - 2 \left( \int 1 \, dx - \int \frac{1}{x^2 + 1} \, dx \right)$$ 8. **Integrate each term:** - $$\int 1 \, dx = x$$ - $$\int \frac{1}{x^2 + 1} \, dx = \arctan x$$ 9. **Combine results:** $$I = x \ln(x^2 + 1) - 2(x - \arctan x) + C = x \ln(x^2 + 1) - 2x + 2 \arctan x + C$$ **Final answer:** $$\boxed{I = x \ln(x^2 + 1) - 2x + 2 \arctan x + C}$$