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$$\int\limits_{1}^{2} x\sqrt{x-1}\;\text{d}x.$$
I've set $u=x-1$ and rearranged the equation to get $x=u+1$. I differentiated $u=x-1$ to $\text{d}u=\text{d}x$.
I've rewritten the integral as $\int_{0}^{1} (u+1)\sqrt{u}\;\text{d}u$ which gives me 2 while the answer is $\frac{16}{15}$. I think I differentiated the substitute equation incorrectly. Could someone explain my mistake to me?
$$\int_0^1(u+1)\sqrt u\ du=\int_0^1(u^{\frac32}+u^{\frac12})du$$
$$=\left(\frac{u^{\frac32+1}}{\frac32+1}+\frac{u^{\frac12+1}}{\frac12+1}\right)\big|_0^1$$
$$=\frac1{\frac52}+\frac1{\frac32}=\frac25+\frac23=\cdots$$
Do $x - 1 = u^2$. If $x = 1$, then $u = 0$. If $x = 2$, then $u = 1$. Otherside, $dx = 2udu$.
$\int_{1}^{2}x\sqrt{x - 1}dx = \int_{0}^{1}(1 + u^2)u\cdot 2udu = 2\int_{0}^{1}(u^2 + u^4)du = \frac{16}{15}$
