_6^16(_ex^2)/(_ex^2+_e(x^2-44x+484)),dx is equal to

Question

$\displaystyle\int_6^{16}\frac{\log_ex^2}{\log_ex^2+\log_e(x^2-44x+484)}\,dx$ is equal to

Accepted Answer

Let $$I = \int\limits_6^{16} {{{{{\log }_e}{x^2}} \over {{{\log }_e}{x^2} + {{\log }_e}({x^2} - 44x + 484)}}dx} $$

$$I = \int\limits_6^{16} {{{{{\log }_e}{x^2}} \over {{{\log }_e}{x^2} + {{\log }_e}({x^2} - 22)}}dx} $$ .... (1)

We know,

$$\int\limits_a^b {f(x)dx}  = \int\limits_a^b {f(a + b - x)} \,dx$$ (king)

So, $$I = \int\limits_6^{16} {{{{{\log }_e}{{(22 - x)}^2}} \over {{{\log }_e}{{(22 - x)}^2} + {{\log }_e}{{(21 - (22 - x))}^2}}}} $$

$$I = \int\limits_6^{16} {{{{{\log }_e}{{(22 - x)}^2}} \over {{{\log }_e}{x^2} + {{\log }_e}{{(22 - x)}^2}}}dx} $$ .... (2)

(1) + (2)

$$2I = \int\limits_6^{16} {1.\,dx}  = 10$$

I = 5

Solutions · 1