$\displaystyle\int \dfrac{e^x}{e^{2x}+1}\,dx$ is: (a) $\log(e^x+1)+c$ (b) $\log(e^{2x}+1)+c$ (c) $\tan^{-1}(e^{2x})+c$ (d) $\tan^{-1}(e^x)+c$

∫ dfracexe2x+1dx is: (a) log(ex+1)+c (b) log(e2x+1)+c (c) tan⁻¹(e2x)+c (d) tan⁻¹(ex)+c

Question 100 of 129

\displaystyle\int \dfrac{e^x}{e^{2x}+1}\,dx

(a)

\log(e^x+1)+c

(b)

\log(e^{2x}+1)+c

(c)

\tan^{-1}(e^{2x})+c

(d)

\tan^{-1}(e^x)+c

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The substitution $u=e^x$ reduces the integral to $\int\frac{du}{u^2+1} = \tan^{-1}u+c$ , giving $\tan^{-1}(e^x)+c$ .

Let $u = e^x$ , so $du = e^x\,dx$ .

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