$\int \frac{3 e^{3 x}}{1+e^{6 x}} d x=\tan ^{-1}\left(e^{3 x}\right)+C$ .

Name: $\int \frac{3 e^{3 x}}{1+e^{6 x}} d x=\tan ^{-1}\left(e^{3 x}\right)+C$ .
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Quizplus · Accepted Answer

The integral can be solved by substituting $u = e^{3x}$, which simplifies the integral to $\int \frac{3u}{1+u^2} du$, leading directly to $\tan^{-1}(u) + C$ or $\tan^{-1}(e^{3x}) + C$.

∫3e3x1+e6xdx=tan⁡−1(e3x)+C\int \frac{3 e^{3 x}}{1+e^{6 x}} d x=\tan ^{-1}\left(e^{3 x}\right)+C∫1+e6x3e3x​dx=tan−1(e3x)+C

$\int \frac{3 e^{3 x}}{1+e^{6 x}} d x=\tan ^{-1}\left(e^{3 x}\right)+C$