Question

If $\int_{}^{}\frac{xe^{x}}{\sqrt{1 + e^{x}}}$dx = ƒ(x) $\sqrt{1 + e^{x}}$– 2 log [g(x)] + c, then find the value of ƒ(x) –

• A. ƒ(x) = 2x – 4, g(x) = $\frac{\sqrt{1 + e^{x}} - 1}{\sqrt{1 + e^{x}} + 1}$
• B. ƒ(x) = 2x + 4, g(x) = $\frac{\sqrt{1 + e^{x}} - 1}{\sqrt{1 + e^{x}} + 1}$
• C. ƒ(x) = 2x – 1, g(x) = $\frac{\sqrt{1 + e^{x}} - 1}{\sqrt{1 + e^{x}} + 1}$
• D. None of these

Answer 1

Answer: (A)

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

Answer 2

Let I = $\int_{}^{}\frac{xe^{x}}{\sqrt{1 + e^{x}}}$dx

Let 1 + e^x = t² Ž e^x dx = 2t dt \ I = 2 $\int_{}^{\int}\log$ (t² – 1) dt

=2 $\left[ \mathrm { t } \log \left( \mathrm { t } ^ { 2 } - 1 \right) - 2 \int \frac { \mathrm { t } ^ { 2 } } { \mathrm { t } ^ { 2 } - 1 } \mathrm { dt } \right]$

=2

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

But $\int_{}^{}\frac{xe^{x}dx}{\sqrt{1 + e^{x}}}$= ƒ(x) $\sqrt{1 + e^{x}}$– 2 log [g(x)] + c

\ ƒ(x) = 2x – 4

And g(x) = $\frac{\sqrt{1 + e^{x}} - 1}{\sqrt{1 + e^{x}} + 1}$.