Question

\[ \int \log (2+x)^{2+x} d x= \]

• A. \(\frac{(2+x)^{2}}{2}\mathrm{log}(\frac{2+x}{\sqrt{\mathrm{e}}})+\mathrm{c}\), where c is the constant of integration
• B. \(\frac{(2+x)^{2}}{2}\mathrm{log}(\frac{2+x}{\mathrm{e}})+\mathrm{c}\), where c is the constant of integration
• C. \(\frac{2+x}{2}\mathrm{log}(\frac{2+x}{\sqrt{\mathrm{e}}})+\mathrm{c}\), where c is the constant of integration
• D. \(\frac{2+x}{2}\mathrm{log}(2+x)\sqrt{\mathrm{e}}+\mathrm{c}\), where c is the constant of integration

Hint:

Use \(\log a^{b} = b\log a\) to get \(\int (2+x)\log(2+x)\,dx\), then substitute \(t = 2+x\) and integrate by parts.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
\(\log (2+x)^{2+x} = (2+x)\log(2+x)\), so the integral becomes \(\int (2+x)\log(2+x)\,dx\).

Step 2: Substitute
Let \(t = 2 + x\), \(dt = dx\). The integral is \(\int t\log t\,dt\).

Step 3: Integrate by parts
\(\int t\log t\,dt = \dfrac{t^2}{2}\log t - \int \dfrac{t^2}{2}\cdot\dfrac{1}{t}\,dt = \dfrac{t^2}{2}\log t - \dfrac{t^2}{4} + c\)

Step 4: Rewrite
\(\dfrac{t^2}{2}\left(\log t - \dfrac{1}{2}\right) + c = \dfrac{(2+x)^2}{2}\log\dfrac{2+x}{\sqrt{e}} + c\)

Final Answer:
The integral equals \(\dfrac{(2+x)^2}{2}\log\dfrac{2+x}{\sqrt{e}} + c\). \[ \boxed{\text{(A) }\dfrac{(2+x)^{2}}{2}\log\left(\dfrac{2+x}{\sqrt{e}}\right)+c} \]