If ∫$$e4x$$−$$1e2xlog$$⁡$$e2x+1e2x$$−$$1dx=t22log$$⁡t−$$t24+u22log$$⁡u−$$u24+C$$ then

Question

If ∫$$e4x$$−$$1e2xlog$$⁡$$e2x+1e2x$$−$$1dx=t22log$$⁡t−$$t24+u22log$$⁡u−$$u24+C$$ then

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The integrand can be written $$ase2x$$−e−$$2xlog$$⁡$$ex+e$$−x−$$e2x$$−e−$$2xlog$$⁡ex−e−$$x=ex+e$$−xex−e−xlog⁡$$ex+e$$−x−$$ex+e$$−xex−e−xlog⁡ex−e−x.Hence the given integrals is equal to∫tlog⁡tdt−∫ulog⁡$$udut=ex+e$$−x,$$u=ex$$−e−$$x=t22log$$⁡t−$$12$$∫$$tdt+u22log$$⁡u−$$12$$∫udu                                                                    $$(integration$$ by $$parts)=t22log$$⁡t−$$t24+u22log$$⁡u−$$u24+Cwhere$$ $$t=ex+e$$−x and  $$u=ex$$−e−x.

If $\int \frac{e^{4 x} - 1}{e^{2 x}} \log (\frac{e^{2 x} + 1}{e^{2 x} - 1}) d x$
$= \frac{t^{2}}{2} \log t - \frac{t^{2}}{4} + \frac{u^{2}}{2} \log u - \frac{u^{2}}{4} + C$ then

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If ∫e4x−1e2xlog⁡e2x+1e2x−1dx=t22log⁡t−t24+u22log⁡u−u24+C then

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If $\int \frac{e^{4 x} - 1}{e^{2 x}} \log (\frac{e^{2 x} + 1}{e^{2 x} - 1}) d x$
$= \frac{t^{2}}{2} \log t - \frac{t^{2}}{4} + \frac{u^{2}}{2} \log u - \frac{u^{2}}{4} + C$ then