Animated Solution for Mathematics - Indefinite Integration: Let $I(x)=\int \frac{x+1}{x(1+x{e}^x{)}^2}dx,x>0$. If ${\lim }_{x\to \infty }I(x)=0$ then $I(1)$ is equal to

Welcome, future engineer. Today, we are going to dismantle a problem that, at first glance, looks like a chaotic mess of exponentials and polynomials. We are looking at the integral $I(x)=\int \frac{x+1}{x(1+x{e}^x{)}^2}dx$.

When you see an expression like this, it is natural to feel a bit overwhelmed. But remember, in JEE Advanced, complexity is often just a mask for elegance. Our job is to peel back that mask.

The Multiplication Trick

The first thing that should catch your eye is the term $(1+x{e}^x{)}^2$ in the denominator. In the world of calculus, whenever you see a function and its derivative potentially lurking nearby, you are on the right track.

Notice the numerator: $x+1$. If we differentiate $x{e}^x$, we get $e^x(1+x)$. This is almost exactly our numerator! The only thing missing is an $e^x$ factor.

So, what do we do? We create it. We multiply the numerator and the denominator by $e^x$. This gives us:

Suddenly, the problem transforms. The numerator is now the perfect differential of $x{e}^x$. This is the 'Aha!' moment that separates the casual student from the master.

The Substitution Dance

Now that we have prepared the ground, let us perform the substitution. Let $t=1+x{e}^x$.

Differentiating both sides with respect to $x$, we get $dt=(e^x\cdot 1+x\cdot e^x)dx$, which simplifies beautifully to $dt=e^x(1+x)dx$. Look at that! Our numerator is exactly $dt$.

What about the $x{e}^x$ term in the denominator? Since $t=1+x{e}^x$, it follows that $x{e}^x=t-1$. Substituting these into our integral, we get:

This is a standard rational function. The terror of the exponential has vanished, replaced by a clean, algebraic structure.

Partial Fraction Decomposition

To solve $\int \frac{1}{(t-1)t^2}dt$, we use partial fractions. We write:

Multiplying by the common denominator $(t-1)t^2$, we get $1=A{t}^2+Bt(t-1)+C(t-1)$. By setting $t=1$, we find $A=1$. By setting $t=0$, we find $C=-1$.

Comparing the coefficients of $t^2$, we see $A+B=0$, which means $B=-1$. Our integral is now:

Integrating term by term, we get $\ln |t-1|-\ln |t|+\frac{1}{t}+C$. Using logarithm properties, this simplifies to $\ln |\frac{t-1}{t}|+\frac{1}{t}+C$.

The Boundary Condition and Final Result

We substitute back $t=1+x{e}^x$ to get $I(x)=\ln |\frac{x{e}^x}{1+x{e}^x}|+\frac{1}{1+x{e}^x}+C$. We are given ${\lim }_{x\to \infty }I(x)=0$.

As $x\to \infty$, the term $\frac{x{e}^x}{1+x{e}^x}$ approaches $1$, so $\ln (1)=0$. The term $\frac{1}{1+x{e}^x}$ approaches $0$. Thus, $0+0+C=0$, implying $C=0$.

Finally, we calculate $I(1)$:

Combining the constants, we get $\frac{1+e+1}{1+e}-\ln (1+e)=\frac{e+2}{e+1}-\ln (e+1)$.

We have arrived at the solution. Remember, the math didn't change; your perspective did. Keep practicing, and keep falling in love with the process.