b. Factorise c. Given that U = x3 − 3x2y + 2xy3 − 3y3, find du/dx (ii) du/dy

Given the substitution $u^2 = x + 1$.
From this, we can express $x$ in terms of $u$:

$$x = u^2 - 1$$

Next, differentiate $u^2 = x + 1$ with respect to $x$ to find $dx$:

$$2u \frac{du}{dx} = 1$$ $$dx = 2u\,du$$

Also, the term $\sqrt{x+1}$ becomes:

$$\sqrt{x+1} = \sqrt{u^2} = u \quad (\text{assuming } u > 0 \text{ for the square root})$$

When $x = 1$:

$$u^2 = 1 + 1 = 2 \implies u = \sqrt{2}$$

When $x = 3$:

$$u^2 = 3 + 1 = 4 \implies u = 2$$

Now, substitute $x$, $\sqrt{x+1}$, and $dx$ into the integral, along with the new limits:

$$\int_{1}^{3} \frac{x}{\sqrt{x+1}}\,dx = \int_{\sqrt{2}}^{2} \frac{u^2 - 1}{u} (2u\,du)$$

$$\int_{\sqrt{2}}^{2} 2(u^2 - 1)\,du$$ $$\int_{\sqrt{2}}^{2} (2u^2 - 2)\,du$$

Now, integrate with respect to $u$:

$$\left[ \frac{2u^3}{3} - 2u \right]_{\sqrt{2}}^{2}$$

Substitute the upper limit and subtract the substitution of the lower limit:

$$\left( \frac{2(2)^3}{3} - 2(2) \right) - \left( \frac{2(\sqrt{2})^3}{3} - 2\sqrt{2} \right)$$ $$\left( \frac{2(8)}{3} - 4 \right) - \left( \frac{2(2\sqrt{2})}{3} - 2\sqrt{2} \right)$$ $$\left( \frac{16}{3} - \frac{12}{3} \right) - \left( \frac{4\sqrt{2}}{3} - \frac{6\sqrt{2}}{3} \right)$$ $$\frac{4}{3} - \left( -\frac{2\sqrt{2}}{3} \right)$$ $$\frac{4}{3} + \frac{2\sqrt{2}}{3}$$ $$\frac{4 + 2\sqrt{2}}{3}$$

The final answer is:

$$\boxed{\frac{4 + 2\sqrt{2}}{3}}$$