JEE Main 2021 — Mathematics Calculus

Let $\displaystyle {\int }_{6}^{16}\frac{{\mathrm{log}}_{e}{x}^{2}dx}{{\mathrm{log}}_{e}{x}^{2}+{\mathrm{log}}_{e}({x}^{2}-44x+484)}$∫616​loge​x2+loge​(x2−44x+484)loge​x2dx​

$${\int }_{6}^{16}\frac{{\mathrm{log}}_{e}{x}^{2}}{{\mathrm{log}}_{e}{x}^{2}+{\mathrm{log}}_{e}{(x-22)}^{2}}dx.....(1)$$∫616​loge​x2+loge​(x−22)2loge​x2​dx.....(1)

We know

$${\int }_{a}^{b}f(x)dx={\int }_{a}^{b}f(a+b-x)dx$$∫ab​f(x)dx=∫ab​f(a+b−x)dx

So, $\displaystyle {\int }_{6}^{16}\frac{{\mathrm{log}}_{e}{(22-x)}^{2}}{{\mathrm{log}}_{e}{(22-x)}^{2}+{\mathrm{log}}_{e}{(22-(22-x))}^{2}}dx$∫616​loge​(22−x)2+loge​(22−(22−x))2loge​(22−x)2​dx

$$I={\int }_{6}^{16}\frac{{\mathrm{log}}_{e}{(22-x)}^{2}}{{\mathrm{log}}_{e}{x}^{2}+{\mathrm{log}}_{e}{(22-x)}^{2}}\mathrm{dx}....(2)$$I=∫616​loge​x2+loge​(22−x)2loge​(22−x)2​dx....(2)

Add (1)&(2)

$$2I={\int }_{6}^{16}dx=10,I=5$$2I=∫616​dx=10,I=5