Question

hello can anyone help me wid this problem in integration ​

Answer 1

Solution :

Let, $\sqrt{1+e^{x}}=z$

Taking differentials, we get

$\frac{1}{2}\frac{e^{x}dx}{\sqrt{1+e^{x}}}=dz$

$\to \frac{e^{x}dx}{\sqrt{1+e^{x}}}=2dz$

Also, $\sqrt{1+e^{x}}=z$

Squaring both sides, we get

$1+e^{x}=z^{2}$

$\to e^{x}=z^{2}-1$

$\to x=log(z^{2}-1)$

Now, $\int \frac{x\:e^{x}}{\sqrt{1+e^{x}}}dx$

$=2\int log(z^{2}-1)dz$

$=2log(z^{2}-1)\int dz-2\int [\frac{d}{dz}\{log(z^{2}-1)\}\times \int dz]dz$

$=2zlog(z^{2}-1)-2\int \frac{2z^{2}dz}{z^{2}-1}$

$=2zlog(z^{2}-1)-4\int \frac{(z^{2}-1)+1}{z^{2}-1}dz$

$=2zlog(z^{2}-1)-4\int dz -4\int \frac{dz}{z^{2}-1}+C$

$=2zlog(z^{2}-1)-4z-\frac{4}{2}log(\frac{z-1}{z+1})+C$

where C is integral constant

$=2x\sqrt{1+e^{x}}-4\sqrt{1+e^{x}}-2log\frac{\sqrt{1+e^{x}}-1}{\sqrt{1+e^{x}}+1}+C$

$=2(x-2)\sqrt{1+e^{x}}-2log\frac{\sqrt{1+e^{x}}-1}{\sqrt{1+e^{x}}+1}+C$

Comparing with the given integral solution, we get

f(x) = 2(x - 2) >> option (d) is correct

g(x) = $\frac{\sqrt{1+e^{x}}-1}{\sqrt{1+e^{x}}+1}$ >> option (b) is correct