\[\frac{1}{e}\log(x^{e} + e^{x}) + c] | MATH - INTEGRATION BY SUBSTITUTION | SolveeIt

$\frac{1}{e}\log(x^{e} + e^{x}) + c$

$\int_{}^{}\frac{\sin x\mspace{6mu} dx}{a^{2} + b^{2}\cos^{2}x} =$

$\log(a^{2} + b^{2}\cos^{2}x) + c$

$\frac{1}{ab}\tan^{- 1}\left( \frac{a\cos x}{b} \right) + c$

None of these

Answer

$\int_{}^{}\frac{\sin x\mspace{6mu} dx}{a^{2} + b^{2}\cos^{2}x} =$

Explanation

$\frac{a^{3x + 3}}{3\log a} + c$

Putting $a^{3x + 3}\log a + c$ , we get

$3a^{3x + 3}\log a + c$

Question: \[\frac{1}{e}\log(x^{e} + e^{x}) + c\]...