Integration by substitution - $\int x^2 \sqrt{x-2} \text{dx}$

Question

$$\int x^2\sqrt{x-2} \, dx,u=x-2$$

Using the given substitution

$u=x-2$

$\text{du}=\text{dx}$

Attempting to express integral in terms of u...

$\int u^2+4x-4\cdot \sqrt{u} \ \text{du}$

This is where I'm stuck - where have I gone wrong?

Answer 1

Given $u=x-2$, we get the following:

• $du = dx$
• $x=u+2$
• $x^2=u^2+4u+4$.

Thus, $$\int x^2 \sqrt{x-2} \,du=\int (u^2+4u+4) u^{\frac 12} \, du = \int u^{\frac 52}+4u^{\frac 32}+4 u^{\frac 12} \, du$$ Now use the anti-power rule, followed by back substitution, to finish it off.