Can you reverse the process?

$\large\begin{aligned}\int_{1}^{\color{#D61F06}a}\frac{x-1}{x+\sqrt{x}}\text{d}x=&\ 4\\\int_{1}^{\color{#D61F06}a}\frac{(\color{#20A900}{\sqrt{x}})^2-\color{teal}1^2}{\sqrt{x}(\sqrt{x}+1)}\text{d}x=&\ 4\\\int_{1}^{\color{#D61F06}a}\frac{(\color{#20A900}{\sqrt{x}}-\color{teal}1)\color{#624F41}(\color{#20A900}{\sqrt{x}}+\color{teal}1\color{#624F41})}{\sqrt{x}\color{#624F41}(\color{#20A900}{\sqrt{x}}+\color{teal}1\color{#624F41})}\text{d}x=&\ 4\\\int_{1}^{\color{#D61F06}a}\frac{\sqrt{x}-1}{\sqrt{x}}\text{d}x=&\ 4\\\int_{1}^{\color{#D61F06}a}1-\dfrac{1}{\sqrt{x}}\text{d}x=&\ 4\\\int_{1}^{\color{#D61F06}a}1-x^{-\frac{1}{2}}\text{d}x=&\ 4\\\left[x-2x^{\frac{1}{2}}\right]_{x=1}^{x=\color{#D61F06}a}=&\ 4\\\color{#D61F06}a-2\sqrt{\color{#D61F06}a}-(1-2)=&\ 4\\\color{#D61F06}a-2\sqrt{\color{#D61F06}a}-3=&\ 0\\(\sqrt{\color{#D61F06}a}-3)(\sqrt{\color{#D61F06}a}+1)=&\ 0\\\sqrt{\color{#D61F06}a}=&\ 3,-1\end{aligned}$

But $\large\sqrt{\color{#D61F06}a}\geq0$ and $\large\color{#D61F06}a>1$ Hence, $\large\sqrt{\color{#D61F06}a}=3\Rightarrow \color{#D61F06}{a=9}$

Thus, $\color{#D61F06}a^2+\color{#D61F06}a+1=\color{#D61F06}9^2+\color{#D61F06}9+1=81+10=\boxed{91}$

Start by making the substitution $x = u^2$ . We do this because it will eliminate the square root in the denominator, which tend to be unmanageable to integrate. With the substitution $x = u^2,$ we have that $dx = 2udu$ , which converts the integral from $\int_{1}^{a}\dfrac{x-1}{x+\sqrt{x}}dx \hspace{.5cm} \text{ to } \hspace{.5cm} \int_{1}^{\sqrt{a}}\dfrac{(u^2-1)}{u^2+u}2udu.$

Simplifying the integrand on the right reduces it to $\int_{1}^{\sqrt{a}}(2u - 2)du = u^2 - 2u \Big|_{1}^{\sqrt{a}} = (a - 2\sqrt{a}) - (-1) = 4$

The only solution to $a - 2\sqrt{a} + 1 = 4$ is $a = 9$ . Therefore, $a^2 + a + 1 = 91$ .

do a little bit of algebraic manipulation:

• $x-1=(\sqrt x-1)(\sqrt x+1)$

• $x + \sqrt(x) = \sqrt(x) ( \sqrt(x) + 1 )$

do some cancelation and evaluate the integral!