∫0ln5exex−1ex+3dx=

∫0ln5exex−1ex+3dxexdx=t2 ⇒exdx=2tdt x→0=t2=0; x=ln5⇒t2=eln5-1 ⇒ t2=4 ⇒t=2=∫02t22tdt(t2+1)+3=2∫02t2t2+4dt=2∫02t2+4-4 t2+4dt=2∫021-4t2+4dt =2t-42tan-1t202 =2(2-2tan-11)-(0-0) =22-2π4 =4-π

∫0ln5exex−1ex+3dx=

∫0ln5exex−1ex+3dxexdx=t2 ⇒exdx=2tdt x→0=t2=0; x=ln5⇒t2=eln5-1 ⇒ t2=4 ⇒t=2=∫02t22tdt(t2+1)+3=2∫02t2t2+4dt=2∫02t2+4-4 t2+4dt=2∫021-4t2+4dt =2t-42tan-1t202 =2(2-2tan-11)-(0-0) =22-2π4 =4-π

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$\int_{0}^{\ln 5} \frac{e^{x} \sqrt{e^{x} - 1}}{e^{x} + 3} d x =$

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$π + 2$

$π - 2$

$4 - π$

$4 + π$

answer is C.

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Detailed Solution

$\int_{0}^{\ln 5} \frac{e^{x} \sqrt{e^{x} - 1}}{e^{x} + 3} d x$

$e^{x} d x = t^{2} \Rightarrow e^{x} d x = 2 t d t x \to 0 = t^{2} = 0; x = \ln 5 \Rightarrow t^{2} = e^{\ln 5} - 1 \Rightarrow t^{2} = 4 \Rightarrow t = 2$

= $\int_{0}^{2} \frac{\sqrt{t^{2}} 2 t d t}{(t^{2} + 1) + 3}$

$= (2) \int_{0}^{2} \frac{t^{2}}{t^{2} + 4} d t$

$= (2) \int_{0}^{2} (\frac{t^{2} + 4 - 4}{t^{2} + 4}) d t$

$= 2 \int_{0}^{2} (1 - \frac{4}{t^{2} + 4}) d t = 2 {[t - \frac{4}{2} \tan^{- 1} \frac{t}{2}]}_{0}^{2} = 2 [(2 - 2 \tan^{- 1} 1) - (0 - 0)] = 2 [2 - 2 (\frac{π}{4})] = 4 - π$