The value of integral ∫0ln 5 exex-1ex+3 dx is ____

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The value of integral $\int_{0}^{\ln 5} \frac{e^{x} \sqrt{e^{x} - 1}}{e^{x} + 3} dx$ is ____

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Detailed Solution

We solve this problem by using the substitution method.
We assume the value inside the square root as a square of some other variable that is

e^{x} - 1 = t^{2}

Then we differentiate the above equation to find the value of

dx

in terms of

dt

Then we convert the given integral into a variable of

t

and also we change the limits of the integration according to the new variable.
We use the standard formula of integration that is

\int \frac{1}{a^{2} + x^{2}} dx = \frac{1}{a} \frac{x}{a}

also

\frac{d}{dx}

(

e^{x}

−1)=

\frac{d}{dx} (t^{2})

⇒

e^{x}

=2t

\frac{dx}{dt}

Now,

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

2 \int_{0}^{2} \frac{t^{2}}{t^{2} + 4} dt

2 \int_{0}^{2} \frac{t^{2} + 4 - 4}{t^{2} + 4} dt

=2(

\int_{0}^{2} 1 dt - 4 \int_{0}^{2} \frac{dt}{t^{2} + 4}

)
=2

[(t - 2 \tan^{- 1} (\frac{t}{2})]

{[(2 - 2 \times \frac{π}{4})]}_{0}^{2}

= 4 - π
Hence (1) is the correct option.

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