치환적분법

(a) Letting $u = x-1$, we find that $x = u+1$ and thus $dx = du$. (b) Using the change of variables in (a), $$\int x \sqrt{x-1} \, dx = \int (u+1) \sqrt{u} \, du\,.$$ (c) Using the fact that $\sqrt{u} = u^{1/2}$ and distributing, we can evaluate the integral as follows: $$\begin{aligned}\int x \sqrt{x-1} \, dx &= \int (u+1) u^{1/2} \, du \\ &= \int \left(u^{3/2} + u^{1/2} \right) \, du \\ &= \frac{2}{5} u^{5/2} + \frac{2}{3} u^{3/2} + C \\ &= \frac{2}{5} (x-1)^{5/2} + \frac{2}{3} (x-1)^{3/2} + C\end{aligned}\,.$$ (d) First, for $\int x^2 \sqrt{x-1} \, dx$, we let $u = x-1$ so that $x = u+1$ and $dx = du$. From there, our work is very similar to what we did in (b): $$\begin{aligned}\int x^2 \sqrt{x-1} \, dx &= \int (u+1)^2 u^{1/2} \, du \\ &= \int \left(u^2 + 2u + 1 \right) u^{1/2} \, du \\ &= \int \left(u^{5/2} + 2u^{3/2} + u^{1/2} \right) \, du \\ &= \frac{2}{7} u^{7/2} + \frac{4}{5} u^{5/2} + \frac{2}{3} u^{3/2} + C \\ &= \frac{2}{7} (x-1)^{7/2} + \frac{4}{5} (x-1)^{5/2} + \frac{2}{3} (x-1)^{3/2} + C\end{aligned}\,.$$ For $\int x \sqrt{x^2 - 1} \, dx$, despite the fact that the integral looks remarkably similar, the situation is quite different. We observe that with $u = x^2 - 1$, it follows $du = 2x \, dx$, so we have a standard function-derivative pair. Noting that $x \, dx = \frac{1}{2} \, du$, it follows that $$\begin{aligned}\int x \sqrt{x^2 - 1} \, dx &= \int \sqrt{u} \cdot \frac{1}{2} \, du \\ &= \frac{1}{2} \cdot \frac{2}{3}u^{3/2} + C \\ &= \frac{1}{3} u^{3/2} + C \\ &= \frac{1}{3} (x^2-1)^{3/2} + C\end{aligned}\,.$$ In the situation where we have $(x-a)$ under a square root and a polynomial outside the square root, the substitution $u = x-a$ will allow us to convert the integrand into an expression we can integrate by changing the form of the radical from $\sqrt{x-a}$ to $\sqrt{u}$. From there we are able to distribute and integrate an expression that is essentially a sum of (fractional) powers of $u$. In the situation where a polynomial of degree higher than $1$ (such as $x^2 - 1$) is under the radical, we likely have to be fortunate enough to have the derivative of said polynomial (up to a constant) outside the radical in order to be able to integrate, as we were in the last example.