Question

$\int_{}^{}\frac{e^{3x} + e^{x}}{e^{4x}–e^{2x} + 1}$ dx =

• A. tan^–1 (e^x – e^–x) + c
• B. tan^–1 (e^x + e^–x) + c
• C. 2 tan^–1 (e^x – e^–x) + c
• D. 2 tan^–1 (e^x + e^–x) + c

Answer 1

Answer: (A)

tan^–1 (e^x – e^–x) + c

Answer 2

Put e^x = t

̃ I = $\int_{}^{}\frac{(t^{2} + 1)dt}{t^{4}–t^{2} + 1}$ = $\int_{}^{}\frac{\left( 1 + \frac{1}{t^{2}} \right)dt}{t^{2}–1 + \frac{1}{t^{2}}}$

put t – $\frac{1}{t}$ = z