Question:
The value of the integral $\int\frac{x+1}{x(1+xe^{x})^{2dx$ is
The value of the integral $\int\frac{x+1}{x(1+xe^{x})^{2dx$ is
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Logic Tip: When you see an expression like $xe^x$ in the denominator, multiplying numerator and denominator by $e^x$ is a classic trick. The derivative of $xe^x$ is $e^x(x+1)$, which conveniently matches the numerator!
Updated On: Apr 28, 2026
- $\log\left|\frac{xe^{x{1+xe^{x\right|+c,$ where c is a constant of integration.
- $\log\left|\frac{xe^{x{1+xe^{x\right|-\frac{1}{1+xe^{x+c,$ where c is a constant of integration.
- $\log|1+xe^{x}|+\frac{1}{1+xe^{x+c,$ where c is a constant of integration.
- $\log\left|\frac{xe^{x{1+xe^{x\right|+\frac{1}{1+xe^{x+c,$ where c is a constant of integration.
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The Correct Option is D
Solution and Explanation
Concept:
This integral requires algebraic manipulation and a suitable substitution. By multiplying the numerator and denominator by $e^x$, we can create a derivative of a combined term in the numerator, allowing for a standard $u$-substitution followed by partial fractions.
Step 1: Manipulate the integrand to prepare for substitution.
Let the given integral be $I$: $$I = \int\frac{x+1}{x(1+xe^{x})^{2dx$$ Multiply the numerator and the denominator by $e^x$: $$I = \int\frac{e^x(x+1)}{xe^x(1+xe^{x})^{2dx$$
Step 2: Apply the substitution method.
Let $t = xe^x$. Differentiating both sides with respect to $x$ using the product rule: $$dt = \left(x \cdot e^x + 1 \cdot e^x\right)dx$$ $$dt = e^x(x+1)dx$$ Substituting this into our integral $I$: $$I = \int\frac{1}{t(1+t)^2}dt$$
Step 3: Decompose the integrand using algebraic manipulation.
We can rewrite the numerator $1$ as $(1+t) - t$: $$I = \int\frac{1+t-t}{t(1+t)^2}dt$$ Split the fraction: $$I = \int\left[\frac{1+t}{t(1+t)^2} - \frac{t}{t(1+t)^2}\right]dt$$ $$I = \int\frac{1}{t(1+t)}dt - \int\frac{1}{(1+t)^2}dt$$ Apply the same trick again for the first integral, rewriting $1$ as $(1+t) - t$: $$I = \int\frac{1+t-t}{t(1+t)}dt - \int\frac{1}{(1+t)^2}dt$$ $$I = \int\left(\frac{1}{t} - \frac{1}{1+t}\right)dt - \int\frac{1}{(1+t)^2}dt$$
Step 4: Integrate and substitute back.
Now, integrate each term with respect to $t$: $$I = \log|t| - \log|1+t| - \left(\frac{-1}{1+t}\right) + c$$ $$I = \log\left|\frac{t}{1+t}\right| + \frac{1}{1+t} + c$$ Substitute $t = xe^x$ back into the expression: $$I = \log\left|\frac{xe^x}{1+xe^x}\right| + \frac{1}{1+xe^x} + c$$
This integral requires algebraic manipulation and a suitable substitution. By multiplying the numerator and denominator by $e^x$, we can create a derivative of a combined term in the numerator, allowing for a standard $u$-substitution followed by partial fractions.
Step 1: Manipulate the integrand to prepare for substitution.
Let the given integral be $I$: $$I = \int\frac{x+1}{x(1+xe^{x})^{2dx$$ Multiply the numerator and the denominator by $e^x$: $$I = \int\frac{e^x(x+1)}{xe^x(1+xe^{x})^{2dx$$
Step 2: Apply the substitution method.
Let $t = xe^x$. Differentiating both sides with respect to $x$ using the product rule: $$dt = \left(x \cdot e^x + 1 \cdot e^x\right)dx$$ $$dt = e^x(x+1)dx$$ Substituting this into our integral $I$: $$I = \int\frac{1}{t(1+t)^2}dt$$
Step 3: Decompose the integrand using algebraic manipulation.
We can rewrite the numerator $1$ as $(1+t) - t$: $$I = \int\frac{1+t-t}{t(1+t)^2}dt$$ Split the fraction: $$I = \int\left[\frac{1+t}{t(1+t)^2} - \frac{t}{t(1+t)^2}\right]dt$$ $$I = \int\frac{1}{t(1+t)}dt - \int\frac{1}{(1+t)^2}dt$$ Apply the same trick again for the first integral, rewriting $1$ as $(1+t) - t$: $$I = \int\frac{1+t-t}{t(1+t)}dt - \int\frac{1}{(1+t)^2}dt$$ $$I = \int\left(\frac{1}{t} - \frac{1}{1+t}\right)dt - \int\frac{1}{(1+t)^2}dt$$
Step 4: Integrate and substitute back.
Now, integrate each term with respect to $t$: $$I = \log|t| - \log|1+t| - \left(\frac{-1}{1+t}\right) + c$$ $$I = \log\left|\frac{t}{1+t}\right| + \frac{1}{1+t} + c$$ Substitute $t = xe^x$ back into the expression: $$I = \log\left|\frac{xe^x}{1+xe^x}\right| + \frac{1}{1+xe^x} + c$$
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