TS EAMCET · Maths · Indefinite Integration
If \(\mathrm{I}_1=\int \frac{e^x}{e^{4 x}+e^{2 x}+1} d x, \mathrm{I}_2=\int \frac{e^{-x}}{e^{-4 x}+e^{-2 x}+1} d x\), then \(\mathrm{I}_2-\mathrm{I}_1=\)
- A \(\frac{1}{2} \log \left(\frac{e^{2 x}-e^{-2 x}+1}{e^{2 x}+e^{-2 x}-1}\right)+\mathrm{c}\)
- B \(\frac{1}{2} \log \left(\frac{e^{2 x}-e^{-2 x}-1}{e^{2 x}+e^{-2 x}+1}\right)+\mathrm{c}\)
- C \(\frac{1}{2} \log \left(\frac{e^{2 x}+e^{-x}+1}{e^{2 x}+e^{-x}-1}\right)+\mathrm{c}\)
- D \(\frac{1}{2} \log \left(\frac{e^x+e^{-x}-1}{e^x+e^{-x}+1}\right)+\mathrm{c}\)
Answer & Solution
Correct Answer
(D) \(\frac{1}{2} \log \left(\frac{e^x+e^{-x}-1}{e^x+e^{-x}+1}\right)+\mathrm{c}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{I}_2 = \int \frac{e^{-x}}{e^{-4x} + e^{-2x} + 1} dx = \int \frac{e^{3x}}{e^{4x} + e^{2x} + 1} dx\) \(\mathrm{I}_2 - \mathrm{I}_1 = \int \frac{e^{3x} - e^x}{e^{4x} + e^{2x} + 1} dx\) Let \(t = e^x\). Then \(dt = e^x dx\).…
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