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: Integration by Parts Let \( u = \log x \implies du = \frac{1}{x} dx \). Let \( dv = \frac{1}{(1+x)^2} dx \implies v = -\frac{1}{1+x} \). Using \( \int u dv = uv - \int v du \): \[ I = -\frac{\log x}{1+x} - \int -\frac{1}{1+x} \cdot \frac{1}{x} dx = -\frac{\log x}{1+x} + \int \frac{1}{x(1+x)} dx \]
Step 2: Partial Fractions \( \frac{1}{x(1+x)} = \frac{1}{x} - \frac{1}{1+x} \). \[ \int \left( \frac{1}{x} - \frac{1}{1+x} \right) dx = \log x - \log(1+x) = \log\left(\frac{x}{1+x}\right) \]
Step 3: Combine terms \[ I = -\frac{\log x}{1+x} + \log\left(\frac{x}{1+x}\right) + c \] This matches Option (D) perfectly.