Question

If I = $\int_{}^{}\frac{1}{e^{x}}$tan^–1 (e^x) dx, then I equals:

• A. – e^–x tan^–1 (e^x) + log (1 + e^2x) + CB
• B. x – e^–x tan ^–1 e^x – $\frac{1}{2}$log (1 + e^x) + C
• C. x – e^–x tan^–1 (e^x) –$\frac{1}{2}$log (1 + e^2x) + C
• D. None of these

Answer 1

Answer: (C)

x – e^–x tan^–1 (e^x) –$\frac{1}{2}$log (1 + e^2x) + C

Answer 2

Put e^x = t to get

I = tan^–1 (t) dt

= (tan^–1 t) –

= –tan^–1 t +

= –tan^–1 t +

= –tan^–1 t + lnt –$\frac { 1 } { 2 }$ln (1+ t²) + C