MCQ
$\int_{}^{} {\frac{{{e^x}}}{{{e^x} + 1}}} \,dx$ =
- A${e^x} + c$
- B$({e^x} + 1) + c$
- ✓$\log ({e^x} + 1) + c$
- Dએકપણ નહિ.
$\therefore $ $\int_{}^{} {\frac{{{e^x}}}{{{e^x} + 1}}dx = \int_{}^{} {\frac{{dt}}{t} = \log t + c = \log ({e^x} + 1) + c} } $.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$f(x) = \left\{ {\begin{array}{*{20}{c}}
{{x^2} + 2mx - 1\,,}&{x \leq 0}\\
{mx - 1\,\,\,\,\,\,\,\,\,\,\,\,\,,}&{x > 0}
\end{array}} \right.$
જો $f (x)$ એક-એક વિધેય હોય તો $'m'$ ની કિમતોનો ગણ મેળવો.