If $\int_0^1 \frac{e^x}{1 + x} dx = m$, then the value of $\int_0^1 \frac{e^x}{(1 + x)^2} dx$ is:

Given $\int_0^1 \frac{e^x}{1 + x} dx = m$

The two integrals are similar - the second has $(1+x)^2$ in the denominator instead of $(1+x)$. Integration by Parts can connect these forms.

For $\int_0^1 \frac{e^x}{(1 + x)^2} dx$, let:

$u = e^x$ → $du = e^x dx$

$dv = \frac{1}{(1+x)^2} dx$ → $v = -\frac{1}{1+x}$

Applying Integration by Parts:

$\int_0^1 \frac{e^x}{(1 + x)^2} dx = \left[e^x \cdot \left(-\frac{1}{1+x}\right)\right]_0^1 - \int_0^1 \left(-\frac{1}{1+x}\right) \cdot e^x dx$

$= \left[-\frac{e^x}{1+x}\right]_0^1 + \int_0^1 \frac{e^x}{1+x} dx$

$= \left[-\frac{e^x}{1+x}\right]_0^1 + m$

Evaluating $\left[-\frac{e^x}{1+x}\right]_0^1$:

At $x = 1$: $-\frac{e^1}{1+1} = -\frac{e}{2}$

At $x = 0$: $-\frac{e^0}{1+0} = -1$

$\left[-\frac{e^x}{1+x}\right]_0^1 = -\frac{e}{2} - (-1)$

$= -\frac{e}{2} + 1$

$\int_0^1 \frac{e^x}{(1 + x)^2} dx = -\frac{e}{2} + 1 + m$

$= m - \frac{e}{2} + 1$

Therefore, the value is $m - \frac{e}{2} + 1$