MHT CET · Maths · Indefinite Integration
\(\int \log (2+x)^{2+x} \mathrm{~d} x=\)
- A \(\frac{(2+x)^2}{2} \log \left(\frac{2+x}{\sqrt{\mathrm{e}}}\right)+\mathrm{c}\), where c is the constant of integration
- B \(\frac{(2+x)^2}{2} \log \left(\frac{2+x}{e}\right)+\mathrm{c}\), where c is the constant of integration
- C \(\frac{2+x}{2} \log \left(\frac{2+x}{\sqrt{\mathrm{e}}}\right)+\mathrm{c}\), where c is the constant of integration
- D \(\frac{2+x}{2} \log (2+x) \sqrt{\mathrm{e}}+\mathrm{c}\), where c is the constant of integration
Answer & Solution
Correct Answer
(A) \(\frac{(2+x)^2}{2} \log \left(\frac{2+x}{\sqrt{\mathrm{e}}}\right)+\mathrm{c}\), where c is the constant of integration
Step-by-step Solution
Detailed explanation
\(\int \log (2+x)^{2+x} \mathrm{~d} x = \int (2+x) \log (2+x) \mathrm{~d} x\) \(\text{Let } u = 2+x, \mathrm{~d} u = \mathrm{d} x \Rightarrow \int u \log u \mathrm{~d} u\)
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