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Miscellaneous Exercise 4 · Q53

Q.Choose the correct option from the given alternatives: ∫0log⁡5exex−1ex+3 dx=\int_0^{\log 5} \dfrac{e^x\sqrt{e^x-1}}{e^x+3}\,dx = (A) 3+2π3+2\pi (B) 4−π4-\pi (C) 2+π2+\pi (D) 4+π4+\pi

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✓ Free question

Let t=ex−1t=\sqrt{e^x-1}, so t2=ex−1t^2=e^x-1, 2t dt=ex dx2t\,dt=e^x\,dx; limits x=0→t=0x=0\to t=0, x=ln⁡5→t=2x=\ln5\to t=2.

∫0ln⁡5exex−1ex+3 dx=∫022t2t2+4 dt=∫02(2−8t2+4)dt=[2t−4tan⁡−1t2]02=4−4tan⁡−11=4−4⋅π4=4−π.\int_0^{\ln5}\frac{e^x\sqrt{e^x-1}}{e^x+3}\,dx=\int_0^2\frac{2t^2}{t^2+4}\,dt=\int_0^2\Big(2-\frac8{t^2+4}\Big)dt=\Big[2t-4\tan^{-1}\frac t2\Big]_0^2=4-4\tan^{-1}1=4-4\cdot\frac\pi4=4-\pi.

✓Final answer

∫0log⁡5exex−1ex+3 dx=4−π\displaystyle\int_0^{\log5}\frac{e^x\sqrt{e^x-1}}{e^x+3}\,dx=4-\pi — option (B)

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