Question

explain me with this calculus problem plz​

Answer 1

Answer:

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

Step-by-step explanation:

We have-

$I = \int {\dfrac{e^{x} }{e^{\frac{x}{2} }+1 } } \, dx$

$Put \ e^{\frac{x}{2} } +1 = t$

$e^{\frac{x}{2} } = (t-1)$

Squaring on both sides

$e^{x} = (t-1)^{2}$

Diff. with respect to 'x'.

$e^{x} dx = 2(t-1)^{2-1}dt$

$e^{x} dx = 2(t-1)dt$

So, $I = \int\ {\dfrac{2(t-1)}{(t-1)}+1 } \, dt$

$I = 2\int\ {\dfrac{t-1}{t} } \, dt$

$I = 2\int\ {[1 - \dfrac{1}{t} } \,] dx$

$I = 2\int\ {1} \, dt  - 2\int\ {\dfrac{1}{t} } \, dt$

$I = 2t - 2logt$

$but \ t = e^{\frac{x}{2} } + 1$

So, $I = 2(e^{\frac{x}{2} }+1) - 2log(e^{\frac{x}{2} }+1)$