Definite integral via u-substitution

Let $u = 3x^2+2x+5$, so $du = (6x+2)\,dx = 2(3x+1)\,dx$, which gives $(3x+1)\,dx = \frac{1}{2}\,du$. When $x=0$, $u=5$; when $x=1$, $u=3+2+5=10$. The integral becomes $\int_5^{10} \frac{1}{\sqrt{u}}\cdot\frac{1}{2}\,du = \frac{1}{2}\int_5^{10} u^{-1/2}\,du$.

Integrating, $\frac{1}{2}\int u^{-1/2}\,du = \frac{1}{2}\cdot 2u^{1/2} = u^{1/2}$. Evaluating from $u=5$ to $u=10$ gives $\sqrt{10}-\sqrt{5}$. Thus the value of the definite integral is $\sqrt{10}-\sqrt{5}$.

The upper bound was x equals one, and substituting into u equals three x squared plus two x plus five gives three plus two plus five, which equals ten. So the upper bound in terms of u is ten.