CUET · MATHS · PYQ PAPER 2023
\(\int \frac{x+e^{2 z}}{x^2+e^{2 x}} d x=\)
- A \(\log \left(x^2+e^{2 x}\right)+C\)
- B \(\frac{1}{2} \log \left(x^2+e^{2 x}\right)+C\)
- C \(\frac{1}{2} \log \left|x^2\right|+C\)
- D \(\log \left|e^{2 x}\right|+C\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{2} \log \left(x^2+e^{2 x}\right)+C\)
Step-by-step Solution
Detailed explanation
Let \(u = x^2 + e^{2x}\). \(du = (2x + 2e^{2x}) dx = 2(x + e^{2x}) dx\). \(\int \frac{x+e^{2 x}}{x^2+e^{2 x}} d x = \int \frac{1}{u} \frac{1}{2} du\). \(= \frac{1}{2} \log|u| + C\). \(= \frac{1}{2} \log(x^2 + e^{2x}) + C\).
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