The integral _2^4 x^2 x^2+(36-12x+x^2),dx is equal to

Question

The integral $\displaystyle\int_2^4 \dfrac{\log x^2}{\log x^2+\log(36-12x+x^2)}\,dx$ is equal to

Accepted Answer

$$I = \int\limits_2^4 {{{\log {x^2}} \over {\log {x^2} + \log \left( {36 - 12x + {x^2}} \right)}}} $$

$$I = \int\limits_2^4 {{{\log {x^2}} \over {\log {x^2} + \log {{\left( {6 - x} \right)}^2}}}} \,\,\,\,\,\,\,\,\,\,...\left( i \right)$$

$$I = \int\limits_2^4 {{{\log {{\left( {6 - x} \right)}^2}} \over {\log {{\left( {6 - x} \right)}^2} + \log {x^2}}}} \,\,\,\,\,\,\,\,...\left( {ii} \right)$$

Adding $(1)$ and $(2)$

$$2I = \int\limits_2^4 {dx = \left[ x \right]_2^4}  = 2 \Rightarrow 1 - 1$$

Solutions · 1