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Definite parametric integral

Let 0<a<b. Evaluate the integral

    \[\mathcal{J} = \int_{a}^{b} \frac{e^{x/a}-e^{b/x}}{\sqrt{abx+ x^3}} \, \mathrm{d}x\]

Solution

The key substitution is x \mapsto \frac{ab}{u}. Applying it we see that

    \begin{align*} \sqrt{abx+ x^3} &\overset{x \mapsto ab/u}{=\! =\! =\! =\! =\!} \sqrt{\frac{a^2b^2}{u} +\frac{a^3b^3}{u^3} } \\ &=\sqrt{\frac{a^2b^2u^2}{u^3} + \frac{a^3b^3}{u^3}} \\ &=\sqrt{\frac{a^2b^2 \left ( u^2+ab \right )}{u^2 \cdot u}} \\ &=\frac{ab}{u} \sqrt{\frac{u^2+ab}{u}} \end{align*}

Thus,

    \begin{align*} \mathcal{J} &= \int_{a}^{b} \frac{e^{x/a}-e^{b/x}}{\sqrt{abx+ x^3}} \, \mathrm{d}x \\ &\!\!\!\!\!\overset{x=ab/u}{=\! =\! =\! =\! =\! =\!} ab\int_{a}^{b} \frac{1}{u^2} \cdot \left ( e^{b/u} - e^{u/a} \right ) \cdot \frac{u}{ab} \cdot \frac{\sqrt{u}}{\sqrt{u^2+ab}} \, \mathrm{d}u \\ &=\int_{a}^{b} \frac{e^{b/u}-e^{u/a}}{\sqrt{abu + u^3}} \, \mathrm{d}u \\ &= - \int_{a}^{b} \frac{e^{u/a}-e^{b/u}}{\sqrt{abu+u^3}} \, \mathrm{d}u \\ &= -\mathcal{J} \end{align*}

Thus , \mathcal{J}=0.

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