MCQ
$\int\limits^\infty_0\frac{1}{1+\text{e}^\text{x}}\text{dx}$ equals:
- ✓$\log2-1$
- B$\log2$
- C$\log4-1$
- D$-\log2$
✓
Answer
Correct option: A.
$\log2-1$
$\int\limits^\frac{\pi^2}{4}_0\frac{\sin\sqrt{\text{x}}}{\sqrt{\text{x}}}\text{dx}$
Let $\sqrt{\text{x}}=\text{t},$ then $\frac{1}{2\sqrt{\text{x}}}\text{dx}=\text{dt}$
when $\text{x}=0,\text{t}=0,\text{x}=\frac{\pi^2}{4},\text{t}=\frac{\pi}{2}$
Therefore the integral becomes
$\int\limits^\frac{\pi}{2}_02\sin\text{t}\text{ dt}$
$=-2\big[\cos\text{t}\big]^\frac{\pi}{2}_0$
$=2$
Let $\sqrt{\text{x}}=\text{t},$ then $\frac{1}{2\sqrt{\text{x}}}\text{dx}=\text{dt}$
when $\text{x}=0,\text{t}=0,\text{x}=\frac{\pi^2}{4},\text{t}=\frac{\pi}{2}$
Therefore the integral becomes
$\int\limits^\frac{\pi}{2}_02\sin\text{t}\text{ dt}$
$=-2\big[\cos\text{t}\big]^\frac{\pi}{2}_0$
$=2$
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