If I=_e^2^e^41xe^1/(( x)^2+1)e^1/(( x)^2+1)+e^1/((6- x)^2+1),dx, then the value of I is:

Question

If $I=\displaystyle\int_{e^2}^{e^4}\frac1x\cdot\frac{e^{1/((\ln x)^2+1)}}{e^{1/((\ln x)^2+1)}+e^{1/((6-\ln x)^2+1)}}\,dx$, then the value of $I$ is:

Accepted Answer

Let $\ln x = t$, $dx/x = dt$:

$$I = \int_2^4 \frac{e^{1/(1+t^2)}}{e^{1/(1+t^2)}+e^{1/(1+(6-t)^2)}}\,dt$$

Applying King's rule ($t\to 6-t$) and adding:

$$2I = \int_2^4 dt = 2 \implies I = 1$$

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