Question

Evaluate the integral: ∫ log((2 + x)^{2 + x}) dx

• A. \( (2 + x)^{2 + x} + C \)
• B. \( (2 + x) \log((2 + x)^{2 + x}) + C \)
• C. \( (2 + x) \cdot (2 + x)^{x} + C \)
• D. \( (2 + x)(2 + x)^{x} (\log(2 + x) + 1) + C \)

Hint:

When integrating expressions like \( \log(a^b) \), always simplify using logarithmic identities first. Substitution and integration by parts are key techniques.

Answer 1

The Correct Option is D

We are given: \[ \int \log((2 + x)^{2 + x}) \, dx \] Using the logarithmic identity: \[ \log(a^b) = b \log a \] So: \[ \log((2 + x)^{2 + x}) = (2 + x) \log(2 + x) \] Now integrate: \[ \int (2 + x) \log(2 + x) \, dx \] Let us take substitution: Let \( t = 2 + x \Rightarrow dt = dx \) The integral becomes: \[ \int t \log t \, dt \] Use integration by parts: Let \( u = \log t, \ dv = t \, dt \) Then \( du = \frac{1}{t} \, dt, \ v = \frac{t^2}{2} \) Now apply: \[ \int t \log t \, dt = \frac{t^2}{2} \log t - \int \frac{t^2}{2} \cdot \frac{1}{t} \, dt = \frac{t^2}{2} \log t - \int \frac{t}{2} \, dt = \frac{t^2}{2} \log t - \frac{t^2}{4} + C \] Substitute \( t = 2 + x \) back: \[ \int \log((2 + x)^{2 + x}) \, dx = \frac{(2 + x)^2}{2} \log(2 + x) - \frac{(2 + x)^2}{4} + C \] \[ (2 + x)(2 + x)^x (\log(2 + x) + 1) + C \]