Q.Evaluate the integral using substitution $\int_{0}^{2}\frac{dx}{x+4-x^2}$

Concept: completing the square, then the standard integral $\int\dfrac{du}{a^2-u^2}$.

Complete the square in the denominator:

With $u=x-\dfrac12$ (so $du=dx$) and $a=\dfrac{\sqrt{17}}{2}$, use

The limits become $u=-\tfrac12$ (at $x=0$) and $u=\tfrac32$ (at $x=2$): …

Step 1 — Complete the square

Factor out the $-1$ on the $x^2$ term and complete the square:

Now $x^2-x-4=\left(x-\tfrac12\right)^2-\tfrac14-4=\left(x-\tfrac12\right)^2-\tfrac{17}{4}$, so

This is $a^2-u^2$ with

Step 2 — Standard integral

The CBSE standard result is

Here $2a=\sqrt{17}$, and multiplying numerator and denominator of $\dfrac{a+u}{a-u}$ by $2$ gives

Step 3 — Change the limits

As $x$ goes from $0$ to $2$, $u=x-\tfrac12$ goes from $-\tfrac12$ to $\tfrac32$. On $[0,2]$ the denominator $x+4-x^2$ stays positive, so the integrand is continuous and the substitution is valid.

Step 4 — Evaluate

At $u=\tfrac32$: $\dfrac{\sqrt{17}+2(3/2)}{\sqrt{17}-2(3/2)}=\dfrac{\sqrt{17}+3}{\sqrt{17}-3}$. …

Method: Complete the square, then match a standard denominator form

For $\displaystyle\int \frac{dx}{\text{quadratic}}$ with no derivative-of-denominator in sight, complete the square to reach $a^2-u^2$ or $u^2+a^2$, then apply the matching standard integral.

Steps

Step 1: Complete the square in the denominator.

Rewrite $\pm x^2+bx+c$ as $\pm[(x+h)^2 \mp a^2]$; a negative $x^2$ coefficient gives the form $a^2-u^2$.

Step 2: Substitute $u=x+h$ (so $du=dx$) and change the limits.

Step 3: Apply the correct standard result.

For a difference of squares, …

Common Mistakes

Mistake 1: Completing the square without handling the negative $x^2$ coefficient.

Why it's wrong: $x+4-x^2=-(x^2-x-4)$; forgetting to factor out the $-1$ flips the form from $a^2-u^2$ to something wrong. Correct approach: rewrite as $\tfrac{17}{4}-(x-\tfrac12)^2$.

Mistake 2: Using the arctan formula for an $a^2-u^2$ denominator. …