1. **State the problem:** Solve the differential equation $$\frac{dy}{dx} = u^2 + 4$$ where $$\int \frac{du}{u^2 + 4} = \int dx$$. 2. **Rewrite the integral:** We have $$\int \frac{du}{u^2 + 4} = \int dx$$ which means $$\int \frac{du}{u^2 + 2^2} = \int dx$$. 3. **Recall the integral formula:** The integral of $$\frac{1}{x^2 + a^2}$$ with respect to $$x$$ is $$\frac{1}{a} \arctan\left(\frac{x}{a}\right) + C$$. 4. **Apply the formula:** Here, $$a = 2$$, so $$\int \frac{du}{u^2 + 4} = \frac{1}{2} \arctan\left(\frac{u}{2}\right) + C$$. 5. **Integrate the right side:** $$\int dx = x + C'$$. 6. **Set the integrals equal:** $$\frac{1}{2} \arctan\left(\frac{u}{2}\right) + C = x + C'$$ which can be simplified to $$\frac{1}{2} \arctan\left(\frac{u}{2}\right) = x + C''$$ where $$C'' = C' - C$$ is a constant. 7. **Solve for $$u$$:** Multiply both sides by 2: $$\arctan\left(\frac{u}{2}\right) = 2x + 2C''$$. 8. **Take the tangent of both sides:** $$\frac{u}{2} = \tan(2x + 2C'')$$ so $$u = 2 \tan(2x + C_1)$$ where $$C_1 = 2C''$$ is an arbitrary constant. **Final answer:** $$u = 2 \tan(2x + C_1)$$