Question

$\int \frac{\log x}{(1+x)^2}dx = $

• A. $\frac{1}{2}\left[\frac{1}{1+x} + \frac{\log x}{(1+x)^2} - \log(x^2+x)\right] + c$
• B. $\frac{1}{2}\left[\frac{1}{1+x} - \frac{\log x}{(1+x)} - \log(1+x^2)\right] + c$
• C. $\frac{1}{2}\left[\frac{1}{1+x} + \frac{\log x}{(1+x)^2} - \log(1+x^2)\right] + c$
• D. $\frac{1}{1+x} + \log x + \log\left[\frac{x}{1+x}\right] + c$
• E. None of these

Hint:

When using integration by parts, the LIATE rule (Logarithmic, Inverse, Algebraic, Trigonometric, Exponential) is a good guideline for choosing the function 'u'. In this case, the logarithmic function $\ln x$ is chosen as 'u' because its derivative is a simpler algebraic function.

Answer 1

Answer: (E)

Step 1: Understanding the Concept
This integral is best solved using the technique of integration by parts. The integrand is a product of a logarithmic function and an algebraic function.
Step 2: Key Formula or Approach
1. Use the integration by parts formula: $\int u dv = uv - \int v du$. 2. Choose $u = \log x$ (or $\ln x$) because its derivative is simpler, and $dv = \frac{1}{(1+x)^2}dx$. 3. Calculate $du$ and $v$. 4. Apply the formula and evaluate the remaining integral, which may require partial fraction decomposition.
Step 3: Detailed Explanation
Let $I = \int \frac{\ln x}{(1+x)^2}dx$. (Assuming $\log x$ is the natural logarithm $\ln x$). 1. Apply Integration by Parts: Let $u = \ln x \implies du = \frac{1}{x}dx$. Let $dv = \frac{1}{(1+x)^2}dx \implies v = \int (1+x)^{-2}dx = -(1+x)^{-1} = -\frac{1}{1+x}$. Using the formula $\int u dv = uv - \int v du$: \[ I = (\ln x)\left(-\frac{1}{1+x}\right) - \int \left(-\frac{1}{1+x}\right)\left(\frac{1}{x}\right)dx \] \[ I = -\frac{\ln x}{1+x} + \int \frac{1}{x(1+x)}dx \] 2. Use Partial Fractions: The second integral requires partial fraction decomposition for the integrand $\frac{1}{x(1+x)}$. \[ \frac{1}{x(1+x)} = \frac{A}{x} + \frac{B}{1+x} \] Multiplying by $x(1+x)$ gives $1 = A(1+x) + Bx$. Setting $x=0$ gives $1 = A(1) \implies A=1$. Setting $x=-1$ gives $1 = B(-1) \implies B=-1$. So, $\frac{1}{x(1+x)} = \frac{1}{x} - \frac{1}{1+x}$. 3. Evaluate the second integral: \[ \int \frac{1}{x(1+x)}dx = \int \left(\frac{1}{x} - \frac{1}{1+x}\right)dx = \ln|x| - \ln|1+x| = \ln\left|\frac{x}{1+x}\right| \] 4. Combine the results: \[ I = -\frac{\ln x}{1+x} + \ln\left(\frac{x}{1+x}\right) + C \] This result can also be written as: \[ I = -\frac{\ln x}{1+x} + \ln x - \ln(1+x) + C \] \[ I = \ln x \left(1 - \frac{1}{1+x}\right) - \ln(1+x) + C = \frac{x\ln x}{1+x} - \ln(1+x) + C \] None of the provided options match this correct result. For example, Option (D) has a positive sign on the first term. Step 4: Final Answer
The correct result of the integration is $-\frac{\ln x}{1+x} + \ln\left(\frac{x}{1+x}\right) + C$. Since this does not match any of the given options, the problem is likely flawed.