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

Let $$I = \int\limits_6^{16} {{{{{\log }_e}{x^2}} \over {{{\log }_e}{x^2} + {{\log }_e}({x^2} - 44x + 484)}}dx} $$<br><br>$$I = \int\limits_6^{16} {{{{{\log }_e}{x^2}} \over {{{\log }_e}{x^2} + {{\log }_e}({x^2} - 22)}}dx} $$ .... (1)<br><br>We know,<br><br>$\int\limits_a^b {f(x)dx} = \int\limits_a^b {f(a + b - x)} \,dx$ (king)<br><br>So, $$I = \int\limits_6^{16} {{{{{\log }_e}{{(22 - x)}^2}} \over {{{\log }_e}{{(22 - x)}^2} + {{\log }_e}{{(21 - (22 - x))}^2}}}} $$<br><br>$$I = \int\limits_6^{16} {{{{{\log }_e}{{(22 - x)}^2}} \over {{{\log }_e}{x^2} + {{\log }_e}{{(22 - x)}^2}}}dx} $$ .... (2)<br><br>(1) + (2)<br><br>$2I = \int\limits_6^{16} {1.\,dx} = 10$<br><br>I = 5