The value of ∫_ 2 ^ 3 (x x^2)/( x^2+ ( 6-x^2)) d x is

Put $x^2=t \Rightarrow x d x=\frac{d t}{2}\therefore \quad I=\int_{\log 2}^{\log 3} \frac{\sin t \cdot \frac{d t}{2}}{\sin t+\sin (\log 6-t)}$ Using, $\quad \int_a^b f(x) d x=\int_a^b f(a+b-x) d x$$\begin{aligned}& I=\frac{1}{2} \int_{\log 2}^{\log 3} \frac{\sin (\log 2+\log 3-t)}{\sin (\log 2+\log 3-t)+\sin } d t \\ & I=\frac{1}{2} \int_{\log 2}^{\log 3} \frac{\sin (\log 6-t)}{\sin (\log 6-t)+\sin (t)} d t \\ & \therefore I=\int_{\log 2}^{\log 3} \frac{\sin (\log 6-t)}{\sin (\log 6-t)+\sin t} d t \quad \ldots(\mathrm{li}) \end{aligned}$$On adding Eqs. (i)$ and (ii), we get 2 I=\frac{1}{2} \int_{\log 2}^{\log 3} \frac{\sin t+\sin (\log 6-t)}{\sin (\log 6-t)+\sin t}$d t$\begin{gathered} \therefore \quad2 I=\frac{1}{2}(t)_{\log 2}^{\log 3}=\frac{1}{2}(\log₃⁻\log 2) \\ \Rightarrow I=\frac{1}{4} \log \left(\frac{3}{2}\right) \end{gathered}