MCQ
$\int_{}^{} {\frac{1}{{{x^2}}}{{(2x + 1)}^3}} dx = $
- ✓$4{x^2} + 12x + 6\log x - \frac{1}{x} + c$
- B$4{x^2} + 12x - 6\log x - \frac{2}{x} + c$
- C$2{x^2} + 8x + 3\log x - \frac{2}{x} + c$
- D$8{x^2} + 6x + 6\log x + \frac{2}{x} + c$
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Consider the following assertions:
$I$. $J>\frac{1}{4}$
$II$. $J<\frac{\pi}{8}$ Then,
Statement $1:$ $f(x)\, \le \,g\,(x)$ for $x$ in $(0,\infty )$
Statement $2:$ $f(x)\, \le \,1$ for $(x)$ in $(0,\infty )$ but $g(x)\,\to \infty$ as $x\,\to \infty$
$f(x)=e^{x-1}-e^{-|x-1|} \text { and } g(x)=\frac{1}{2}\left(e^{x-1}+e^{1-x}\right) \text {. }$ Then the area of the region in the first quadrant bounded by the curves $y=f(x), y=g(x)$ and $x=0$ is