MHT CET · Maths · Indefinite Integration
\(\int(1+x) \log x \mathrm{~d} x=\)
- A \(\left(x+\frac{x^{2}}{2}\right) \log x+\left(x-\frac{x^{2}}{4}\right)+C\)
- B \(\left(x+\frac{x^{2}}{2}\right) \log x-\left(x+\frac{x^{2}}{4}\right)+C\)
- C \(\left(x+\frac{x^{2}}{2}\right) \log x-\left(x-\frac{x^{2}}{4}\right)+C\)
- D \(\left(x+\frac{x^{2}}{2}\right) \log x+\left(x+\frac{x^{2}}{4}\right)+C\)
Answer & Solution
Correct Answer
(B) \(\left(x+\frac{x^{2}}{2}\right) \log x-\left(x+\frac{x^{2}}{4}\right)+C\)
Step-by-step Solution
Detailed explanation
Let \(I=\int(1+x) \log x d x\)
Put \(\log x=t \quad \Rightarrow x=e^{t} \Rightarrow d x=e^{t} d t\)
\(\therefore I =\int\left(1+e^{t}\right) t \cdot e^{t} d t=\int\left(t e^{t}+t e^{2 t}\right) d t \)
\( =\int t e^{t} d t+\int t e^{2 t} d t=\left[t e^{t}-\int e^{t} d t\right]\) \(+\left[t \frac{e^{2 t}}{2}-\int \frac{e^{2 t}}{2} d t\right] \)
\( =t e^{t}-e^{t}+t \frac{e^{2 t}}{2}-\frac{e^{2 t}}{4}+C=x \log x-\) \(x+\frac{1}{2}(\log x) \cdot x^{2}-\frac{x^{2}}{4}+C \)
\( =\left(x+\frac{x^{2}}{2}\right) \log x-\left(x+\frac{x^{2}}{4}\right)+C\)
Put \(\log x=t \quad \Rightarrow x=e^{t} \Rightarrow d x=e^{t} d t\)
\(\therefore I =\int\left(1+e^{t}\right) t \cdot e^{t} d t=\int\left(t e^{t}+t e^{2 t}\right) d t \)
\( =\int t e^{t} d t+\int t e^{2 t} d t=\left[t e^{t}-\int e^{t} d t\right]\) \(+\left[t \frac{e^{2 t}}{2}-\int \frac{e^{2 t}}{2} d t\right] \)
\( =t e^{t}-e^{t}+t \frac{e^{2 t}}{2}-\frac{e^{2 t}}{4}+C=x \log x-\) \(x+\frac{1}{2}(\log x) \cdot x^{2}-\frac{x^{2}}{4}+C \)
\( =\left(x+\frac{x^{2}}{2}\right) \log x-\left(x+\frac{x^{2}}{4}\right)+C\)
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