Finding the integral $\int\frac{e^{x}}{e^{2x}+1}$

As Jean-Claude Arbaut said, you didn't screw up. We have the identity $$\tan^{-1}(\frac{1}{y})=\frac{\pi}{2}-\tan^{-1}(y).$$ So it follows that $$ -\tan^{-1}(1/e^x)=\tan^{-1}(e^x)-\pi/2=\tan^{-1}(e^x)+c. $$


Your answer is right .. Here is a easier approach

$$e^x=\tan z$$ $$\int \frac{e^{x}}{e^{2x}+1}dx$$ $$\int \frac{\sec^2 z}{\tan^2 z+1}dz$$ $$=z+c=\tan^{-1}e^x+c$$


You have been messing a little with the signs. $$\int\frac{e^x\,dx}{e^{2x}+1}=\int\frac{d(e^x)}{(e^x)^2+1}=\arctan(e^x)+c.$$

Tags:

Integration

Calculus

Indefinite Integrals

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