ExamBro
ExamBro
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 \(\mathrm{I}_2-\mathrm{I}_1=\)

  1. A \(\frac{1}{2} \log \left(\frac{e^{2 x}-e^{-2 x}+1}{e^{2 x}+e^{-2 x}-1}\right)+\mathrm{c}\)
  2. B \(\frac{1}{2} \log \left(\frac{e^{2 x}-e^{-2 x}-1}{e^{2 x}+e^{-2 x}+1}\right)+\mathrm{c}\)
  3. C \(\frac{1}{2} \log \left(\frac{e^{2 x}+e^{-x}+1}{e^{2 x}+e^{-x}-1}\right)+\mathrm{c}\)
  4. D \(\frac{1}{2} \log \left(\frac{e^x+e^{-x}-1}{e^x+e^{-x}+1}\right)+\mathrm{c}\)
Verified Solution

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 \(dt = e^x dx\).…