For $x≠1$, if $\int\frac{xe^xdx}{(1+x)^2}=\frac{ae^x}{(1+x)^b}+c$, where $a, b$ are fixed numbers and $c$ is the integration constant, then $a + b$ is equal to

2

The correct answer is Option (3) → 2

Given:

$\displaystyle \int \frac{x e^x}{(1 + x)^2} dx = \frac{a e^x}{(1 + x)^b} + c$

Use substitution: let $u = 1 + x \Rightarrow du = dx$, and $x = u - 1$

Then the integrand becomes:

$\displaystyle \int \frac{(u - 1)e^{u - 1}}{u^2} du$

$= e^{-1} \int \frac{(u - 1)e^u}{u^2} du$

Use integration by parts: Let $I = \int \frac{(u - 1)e^u}{u^2} du$

Split numerator: $I = \int \left( \frac{e^u}{u} - \frac{e^u}{u^2} \right) du$

So the integral becomes:

$\int \frac{x e^x}{(1 + x)^2} dx = e^{-1} \left( \int \frac{e^u}{u} du - \int \frac{e^u}{u^2} du \right)$

This expression is complicated, but the original integral is clearly designed to match the derivative of the RHS:

Let us directly differentiate RHS: $f(x) = \frac{a e^x}{(1 + x)^b}$

$f'(x) = a \cdot \frac{e^x (1 + x)^b - b(1 + x)^{b - 1} \cdot e^x}{(1 + x)^{2b}}$

$= a e^x \cdot \frac{(1 + x)^b - b(1 + x)^{b - 1}}{(1 + x)^{2b}}$

$= a e^x \cdot \frac{(1 + x) - b}{(1 + x)^{b + 1}}$

We want this to match: $\frac{x e^x}{(1 + x)^2}$

So set: $\displaystyle a e^x \cdot \frac{(1 + x) - b}{(1 + x)^{b + 1}} = \frac{x e^x}{(1 + x)^2}$

Cancel $e^x$:

$a \cdot \frac{(1 + x) - b}{(1 + x)^{b + 1}} = \frac{x}{(1 + x)^2}$

Now equate numerators and denominators:

Denominator: $b + 1 = 2 \Rightarrow b = 1$

Numerator: $a((1 + x) - 1) = a x = x \Rightarrow a = 1$

So $a = 1$, $b = 1$ ⇒ $a + b = {2}$