\int\limits_6^16 \log _ex^2 \over \log _ex^2 + \log _e(x^2 - 44x + 484)dx is equal to :

Definite Integration PYQ — JEE Main 2021 · 27 Aug · Shift 1 · Q32 with Solution

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Correct answer: C

We need to evaluate

$I=\int_6^{16} \frac{\ln(x^2)}{\ln(x^2)+\ln(x^2-44x+484)}\,dx.$

Observe that

$x^2-44x+484=(x-22)^2.$

So the integral becomes

$I=\int_6^{16} \frac{\ln(x^2)}{\ln(x^2)+\ln((x-22)^2)}\,dx.$

Using $\ln(a^2)=2\ln|a|$ ,

$\ln(x^2)=2\ln|x|, \qquad \ln((x-22)^2)=2\ln|x-22|.$

On the interval $[6,16]$ , we have $x>0$ and $x-22<0$ , so

$|x|=x, \qquad |x-22|=22-x.$

Hence

=\int_6^{16} \frac{\ln x}{\ln x+\ln(22-x)}\,dx.$$ Thus $$I=\int_6^{16} \frac{\ln x}{\ln(x(22-x))}\,dx.$$ 3. **Use symmetry** Let $$f(x)=\frac{\ln x}{\ln x+\ln(22-x)}.$$ Now consider $f(22-x)$: $$f(22-x)=\frac{\ln(22-x)}{\ln(22-x)+\ln x}.$$ Therefore, $$f(x)+f(22-x)=\frac{\ln x}{\ln x+\ln(22-x)}+\frac{\ln(22-x)}{\ln x+\ln(22-x)}=1.$$ Also, the interval $[6,16]$ is symmetric about $11$, since under the substitution $x\mapsto 22-x$: - $x=6 \mapsto 16$ - $x=16 \mapsto 6$ So, $$I=\int_6^{16} f(x)\,dx.$$ Using the symmetry property, $$I=\int_6^{16} f(22-x)\,dx.$$ Adding the two expressions, $$2I=\int_6^{16} \left[f(x)+f(22-x)\right]dx=\int_6^{16} 1\,dx=16-6=10.$$ Hence, $$I=\frac{10}{2}=5.$$ 4. **Evaluate options** The value of the integral is $$\boxed{5}.$$ So the correct option is **C**.