Evaluate the following integrals: (i) ∫ cos(3x+2)dx (ii) ∫ x3 ln(x)dx (iii) ∫ (3x3+5)/x4 dx

Let $u = 3x+2$.
Then, differentiate $u$ with respect to $x$:

$$\frac{du}{dx} = 3$$ $$du = 3\,dx$$ $$dx = \frac{1}{3}\,du$$

$$\int \cos(u) \left(\frac{1}{3}\right)\,du$$ $$= \frac{1}{3} \int \cos(u)\,du$$

The integral of $\cos(u)$ is $\sin(u)$.

$$= \frac{1}{3} \sin(u) + C$$

$$= \frac{1}{3} \sin(3x+2) + C$$

The final answer is $\boxed{\frac{1}{3} \sin(3x+2) + C}$.

Choose $u = \ln x$ and $dv = x^3\,dx$.
Then, find $du$ and $v$:

$$du = \frac{1}{x}\,dx$$ $$v = \int x^3\,dx = \frac{x^{3+1}}{3+1} = \frac{x^4}{4}$$

$$\int x^3 \ln x\,dx = (\ln x)\left(\frac{x^4}{4}\right) - \int \left(\frac{x^4}{4}\right)\left(\frac{1}{x}\right)\,dx$$ $$= \frac{x^4}{4} \ln x - \int \frac{x^3}{4}\,dx$$

$$= \frac{x^4}{4} \ln x - \frac{1}{4} \int x^3\,dx$$ $$= \frac{x^4}{4} \ln x - \frac{1}{4} \left(\frac{x^4}{4}\right) + C$$ $$= \frac{x^4}{4} \ln x - \frac{x^4}{16} + C$$

$$= \frac{x^4}{4} \left(\ln x - \frac{1}{4}\right) + C$$

The final answer is $\boxed{\frac{x^4}{4} \ln x - \frac{x^4}{16} + C}$.

Let $u = x^2+3$.
Then, differentiate $u$ with respect to $x$:

$$\frac{du}{dx} = 2x$$ $$du = 2x\,dx$$

Notice that $6x\,dx = 3 \cdot (2x\,dx)$.
So, $6x\,dx = 3\,du$.
The integral becomes:

$$\int \frac{3\,du}{u}$$ $$= 3 \int \frac{1}{u}\,du$$

The integral of $\frac{1}{u}$ is $\ln|u|$.

$$= 3 \ln|u| + C$$

Since $x^2+3$ is always positive, we can remove the absolute value.

$$= 3 \ln(x^2+3) + C$$

The final answer is $\boxed{3 \ln(x^2+3) + C}$.