We can write above integrand as:

Considering integrand (A)

Put e^x+1 = t

Differentiating w.r.t x we get,

e^xdx = dt

Substituting values we get

Substituting the value of t we get,

--(i)

Considering integrand (B)

We can write above integral as

(1) (2)

Considering first integral:

Since the numerator and denominator are exactly same, our integrand simplifies to 1 and integrand becomes:

⇒ ∫ dx

⇒ x

--- (3)

Considering second integral:

Let u = 1 + e^x, du = e^xdx

Apply u – substitution:

Replacing the value of u we get,

---(4)

From (3) and (4) we get,

--(ii)

From (i) and (ii) we get,