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Question: If \(\int_{}^{}\frac{xe^{x}}{\sqrt{(1 + e^{x})}}\)dx = ƒ(x) \(\sqrt{(1 + e^{x})}\) – 2 log g(x) + C,...

If ∫xex(1+ex)\int_{}^{}\frac{xe^{x}}{\sqrt{(1 + e^{x})}}dx = ƒ(x) (1+ex)\sqrt{(1 + e^{x})} – 2 log g(x) + C, then –

A

ƒ(x) = x – 1

B

g (x) = 1+ex−11+ex+1\frac{\sqrt{1 + e^{x}} - 1}{\sqrt{1 + e^{x}} + 1}

C

g(x) = 1+ex+11+ex−1\frac{\sqrt{1 + e^{x}} + 1}{\sqrt{1 + e^{x}} - 1}

D

g(x) = 1+ex+11+ex−1\frac{\sqrt{1 + e^{x}} + 1}{\sqrt{1 + e^{x}} - 1}

Answer

g (x) = 1+ex−11+ex+1\frac{\sqrt{1 + e^{x}} - 1}{\sqrt{1 + e^{x}} + 1}

Explanation

Solution

Let I = ∫xex(1+ex)dx\int_{}^{}{\frac{xe^{x}}{\sqrt{(1 + e^{x})}}dx}

=

I II

= x . 2(1+ex)\sqrt{(1 + e^{x})} – ∫1.2(1+ex)\int_{}^{}{1.2\sqrt{(1 + e^{x})}} dx

= 2x (1+ex)\sqrt{(1 + e^{x})} – 2∫(1+ex)\int_{}^{}\sqrt{(1 + e^{x})}dx

in second integral

put 1 + ex = t2

\ dx = 2tdtt2−1\frac{2tdt}{t^{2} - 1}

then = 2x (1+ex)\sqrt{(1 + e^{x})} – 4 ∫t2−1+1(t2−1)\int_{}^{}\frac{t^{2} - 1 + 1}{(t^{2} - 1)}dt

= 2x(1+ex)\sqrt{(1 + e^{x})} – 4(1+1t2−1)\left( 1 + \frac{1}{t^{2} - 1} \right)dt

= 2x(1+ex)\sqrt{(1 + e^{x})}– 4{t+12log⁡ (t−1t+1)] \left\{ t + \frac{1}{2}\log\left. \ \left( \frac{t - 1}{t + 1} \right) \right\rbrack \right.\ + c

= 2x1+ex\sqrt{1 + e^{x}}– 4(1+ex)\sqrt{(1 + e^{x})}– 2 log (1+ex−11+ex+1)\left( \frac{\sqrt{1 + e^{x}} - 1}{\sqrt{1 + e^{x}} + 1} \right) + c

On comparing

ƒ(x) = 2x – 4, g(x) = 1+ex−11+ex+1\frac{\sqrt{1 + e^{x}} - 1}{\sqrt{1 + e^{x}} + 1}.