Q.Complete the following activity:
$\displaystyle\int_0^2 \dfrac{dx}{4 + x - x^2}$
$= \displaystyle\int_0^2 \dfrac{dx}{-x^2 + \square + \square}$
$= \displaystyle\int_0^2 \dfrac{dx}{-x^2 + x + \frac{1}{4} - \square + 4}$
$= \displaystyle\int_0^2 \dfrac{dx}{\left(x - \frac{1}{2}\right)^2 - (\square)^2}$
$= \dfrac{1}{\sqrt{17}} \log\left(\dfrac{20 + 4\sqrt{17}}{20 - 4\sqrt{17}}\right)$

Rewrite the denominator $4+x-x^2$, complete the square as $\tfrac{17}{4}-\left(x-\tfrac12\right)^2 = \left(\tfrac{\sqrt{17}}{2}\right)^2-\left(x-\tfrac12\right)^2$, apply the standard integral, and evaluate from $0$ to $2$. Blanks: $x,\ 4,\ \tfrac14,\ \tfrac{\sqrt{17}}{2}$.

We fill the boxes step by step.

Line 1 — rewrite $4 + x - x^2$ in descending powers of $x$: $\;-x^2 + x + 4$. So the first box is $x$ and the second is $4$:

$\int_0^2 \frac{dx}{4+x-x^2} = \int_0^2 \frac{dx}{-x^2 + x + 4}.$

Line 2 — complete the square. Take $-1$ common from the $x$-terms: $-\left(x^2 - x\right) + 4$; adding and subtracting $\tfrac14$ inside gives $-x^2 + x + \tfrac14 - \tfrac14 + 4$, so the box is $\tfrac14$:

$= \int_0^2 \frac{dx}{-x^2 + x + \tfrac14 - \tfrac14 + 4}.$

Line 3 — group the perfect square. Since $-x^2 + x + \tfrac14 = -\left(x-\tfrac12\right)^2$ and $-\tfrac14+4=\tfrac{15}{4}$... more directly, $-x^2+x+4 = \tfrac{17}{4} - \left(x-\tfrac12\right)^2$. Writing $\tfrac{17}{4}=\left(\tfrac{\sqrt{17}}{2}\right)^2$, the denominator takes the standard form (the box is $\tfrac{\sqrt{17}}{2}$):

$= \int_0^2 \frac{dx}{\left(\tfrac{\sqrt{17}}{2}\right)^2 - \left(x-\tfrac12\right)^2}.$

(Note: the printed activity shows this line with the squares in reversed order; the mathematically correct completed square is $\left(\tfrac{\sqrt{17}}{2}\right)^2 - \left(x-\tfrac12\right)^2$, which is what yields the given final answer.)

Line 4 — apply the standard integral $\displaystyle\int\frac{du}{a^2-u^2}=\frac{1}{2a}\log\left|\frac{a+u}{a-u}\right|$ with $a=\tfrac{\sqrt{17}}{2}$ and $u = x-\tfrac12$ (so $2a=\sqrt{17}$):

$= \frac{1}{\sqrt{17}}\left[\log\left|\frac{\tfrac{\sqrt{17}}{2} + \left(x-\tfrac12\right)}{\tfrac{\sqrt{17}}{2} - \left(x-\tfrac12\right)}\right|\right]_0^2.$

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