Let's break down and solve these integrals step by step using substitution methods.

Q1) $\int \frac{\sec(\ln(x)) \tan(\ln(x))}{x} \, dx$

Here, a substitution of $u = \ln(x)$ will simplify the integral.

• $du = \frac{1}{x} dx$, so $dx = x \, du$.

The integral becomes: $\int \sec(u) \tan(u) \, du$ The derivative of $\sec(u)$ is $\sec(u) \tan(u)$, so the integral is: $\sec(u) + C = \sec(\ln(x)) + C$ Thus, the solution to Q1 is: $\boxed{\sec(\ln(x)) + C}$

Q2) $\int \sqrt{e^x} \, e^x \, dx$

We simplify the expression inside the integral: $\sqrt{e^x} = e^{x/2}$ Thus, the integral becomes: $\int e^{x/2} \cdot e^x \, dx = \int e^{3x/2} \, dx$

Now, use substitution: let $u = \frac{3x}{2}$, so $du = \frac{3}{2} dx$, or $dx = \frac{2}{3} du$.

The integral becomes: $\frac{2}{3} \int e^u \, du = \frac{2}{3} e^u + C$ Substitute back $u = \frac{3x}{2}$: $\boxed{\frac{2}{3} e^{3x/2} + C}$

Q3) $\int \frac{e^{\sqrt{x}}}{2\sqrt{x}} \, dx$

Here, use the substitution $u = \sqrt{x}$, so that $du = \frac{1}{2\sqrt{x}} dx$, or equivalently $dx = 2\sqrt{x} \, du$.

The integral simplifies to: $\int e^u \, du = e^u + C$ Substitute back $u = \sqrt{x}$: $\boxed{e^{\sqrt{x}} + C}$

Q4) $\int \sqrt{4x + 3} \, dx$

Let $u = 4x + 3$, then $du = 4 \, dx$, or $dx = \frac{du}{4}$.

The integral becomes: $\frac{1}{4} \int \sqrt{u} \, du = \frac{1}{4} \int u^{1/2} \, du$ Integrating $u^{1/2}$: $\frac{1}{4} \cdot \frac{2}{3} u^{3/2} + C = \frac{1}{6} (4x + 3)^{3/2} + C$ Thus, the solution is: $\boxed{\frac{1}{6} (4x + 3)^{3/2} + C}$

Would you like further clarification on any of these integrals?

Follow-up questions:

1. What are the key steps in applying substitution for each problem?
2. How does changing the variable simplify each integral?
3. Why does $\sec(u)$ differentiate to $\sec(u)\tan(u)$?
4. Could these problems be solved with other methods like integration by parts?
5. What are other common integrals involving $\sec$ and $\tan$?

Tip:

When performing substitution, always remember to adjust both the differential and the bounds (if definite integrals) accordingly.