AP EAMCET · Maths · Definite Integration
\(\int_0^1(1+x) \log (1+x) d x=\)
- A \(\frac{-3}{4}+\log 2\)
- B \(\frac{3}{4}+2 \log 2\)
- C \(2 \log 2\)
- D \(\frac{-3}{4}+2 \log 2\)
Answer & Solution
Correct Answer
(D) \(\frac{-3}{4}+2 \log 2\)
Step-by-step Solution
Detailed explanation
\text { } \begin{aligned} I & =\int_0^1(1+x) \log (1+x) d x \\ & =\left[(\log (1+x))\left(x+\frac{x^2}{2}\right)\right]_0^1-\int_0^1 \frac{\left(x+\frac{x^2}{2}\right)}{1+x} d x \\ & =\frac{3}{2} \log _e 2-\frac{1}{2} \int_0^1\left[(x+1)-\frac{1}{1+x}\right] d x \\ &…
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