a) Integrate : dxx^2 + 9 . [2] b) Solve: dydx = 1 y2x+1 . [3]

a) Integrate: $\int \frac{dx}{x^2+9}$

Step 1: Identify the standard form. The integral is of the form $\int \frac{dx}{x^2+a^2}$, where $a^2=9⟹a=3$.

Step 2: Apply the standard formula. $$\int \frac{dx}{x^2+a^2}=\frac{1}{a}{\tan }^{-1}\left(\frac{x}{a}\right)+C$$

Step 3: Substitute $a=3$. $$\int \frac{dx}{x^2+9}=\frac{1}{3}{\tan }^{-1}\left(\frac{x}{3}\right)+C$$

Final Answer: $$\frac{1}{3}{\tan }^{-1}\left(\frac{x}{3}\right)+C$$

---

b) Solve: $\frac{dy}{dx}=\frac{1-y}{2x+1}$

Step 1: Separate the variables. Rearrange the differential equation to group terms involving $y$ on the left and terms involving $x$ on the right: $$\frac{dy}{1-y}=\frac{dx}{2x+1}$$

Step 2: Integrate both sides. $$\int \frac{dy}{1-y}=\int \frac{dx}{2x+1}$$

Step 3: Evaluate the integrals. For the LHS, let $u=1-y$, then $du=-dy$. $$\int \frac{-du}{u}=-\ln |u|=-\ln |1-y|$$

For the RHS, let $v=2x+1$, then $dv=2dx⟹dx=\frac{dv}{2}$. $$\int \frac{1}{v}\cdot \frac{dv}{2}=\frac{1}{2}\ln |v|=\frac{1}{2}\ln |2x+1|$$

Combining these and adding the constant of integration $C$: $$-\ln |1-y|=\frac{1}{2}\ln |2x+1|+C$$

Step 4: Simplify the equation. Multiply by $-1$: $$\ln |1-y|=-\frac{1}{2}\ln |2x+1|-C$$

Use logarithmic properties ($a\ln b=\ln b^a$ and $-\ln b=\ln b^{-1}$): $$\ln |1-y|=\ln \left((2x+1{)}^{-1/2}\right)+C_1$$ (Where $C_1=-C$ is an arbitrary constant)

Exponentiate both sides: $$|1-y|=e^{C_1}\cdot \frac{1}{\sqrt{2x+1}}$$

Let $A=\pm e^{C_1}$ be an arbitrary non-zero constant. $$1-y=\frac{A}{\sqrt{2x+1}}$$

Step 5: Solve for $y$. $$y=1-\frac{A}{\sqrt{2x+1}}$$

Final Answer: $$y=1-\frac{A}{\sqrt{2x+1}}$$