61.

$\int \frac{{\mathrm{x}}^2 - 1}{{\mathrm{x}}^4 + 3{\mathrm{x}}^2 + 1}\mathrm{dx}(\mathrm{x} > 0) \mathrm{is}$

• ${\tan }^{-1}\left(\mathrm{x} + \frac{1}{\mathrm{x}}\right) +\hspace{0.17em}\mathrm{C}$

• ${\tan }^{-1}\left(\mathrm{x} - \frac{1}{\mathrm{x}}\right) +\hspace{0.17em}\mathrm{C}$

• ${\log }_{\mathrm{e}}\left|\frac{\mathrm{x} + {\displaystyle \frac{1}{\mathrm{x}}} - 1}{\mathrm{x} + \frac{1}{\mathrm{x}} + 1}\right| + \mathrm{C}$

• ${\log }_{\mathrm{e}}\left|\frac{\mathrm{x} - {\displaystyle \frac{1}{\mathrm{x}}} - 1}{\mathrm{x} - \frac{1}{\mathrm{x}} + 1}\right| + \mathrm{C}$

Let I = ${\int }_{10}^{19}\frac{\sin \left(\mathrm{x}\right)}{1 + {\mathrm{x}}^8}d\mathrm{x}$, then

• $\left|\mathrm{I}\right| < {10}^{- 9}$

• $\left|\mathrm{I}\right| < {10}^{- 7}$

• $\left|\mathrm{I}\right| < {10}^{- 5}$

• $\left|\mathrm{I}\right| > {10}^{- 7}$

Let I₁ = ${\int }_0^{\mathrm{n}}\left[\mathrm{x}\right]d\mathrm{x}$ and I₂ = ${\int }_0^{\mathrm{n}}\left\{\mathrm{x}\right\}d\mathrm{x}$, where [x] and {x} are integral and fractional parts of x and n $\in$ N - {1}. Then, I₁/I₂ is equal to

• $\frac{1}{\mathrm{n} - 1}$

• $\frac{1}{\mathrm{n}}$

• n

• n - 1

The value of $\underset{\mathrm{n}\to \infty }{\lim }\left[\frac{\mathrm{n}}{{\mathrm{n}}^2 + 1^2} +\frac{{\displaystyle \mathrm{n}}}{{\displaystyle {\mathrm{n}}^2 + 2^2}} + ... +\hspace{0.17em}\frac{1}{2\mathrm{n}}\right]$ is

• $\frac{\mathrm{n\pi }}{4}$

• $\frac{\mathrm{\pi }}{4}$

• $\frac{\mathrm{\pi }}{4\mathrm{n}}$

• $\frac{\mathrm{\pi }}{2\mathrm{n}}$

The value of ${\int }_0^{100}{\mathrm{e}}^{{\mathrm{x}}^2}d\mathrm{x}$

• is less than 1

• is greater than 1

• is less than or equal to 1

• lies in the closed interval [1, e]

${\int }_0^{100}{\mathrm{e}}^{\mathrm{x} - \left[\mathrm{x}\right]}d\mathrm{x}$ is equal to

• $\frac{{\mathrm{e}}^{100} - 1}{100}$

• $\frac{{\mathrm{e}}^{100} - 1}{\mathrm{e} - 1}$

• $100(\mathrm{e} - 1)$

• $\frac{\mathrm{e} - 1}{100}$

Solution of ${\left(\mathrm{x} + \mathrm{y}\right)}^2\frac{d\mathrm{y}}{d\mathrm{x}} = {\mathrm{a}}^2$ ( 'a' being a constant) is

• $\frac{\left(\mathrm{x} + \mathrm{y}\right)}{\mathrm{a}} = \tan \left(\frac{\mathrm{y} + \mathrm{C}}{\mathrm{a}}\right)$, C is an arbitrary

• $\mathrm{xy} = \mathrm{atan}\left(\mathrm{Cx}\right)$, C is an arbitrary

• $\frac{\mathrm{x}}{\mathrm{a}} = \tan \left(\frac{\mathrm{y}}{\mathrm{C}}\right)$, C is an arbitrary

• $\mathrm{xy} = \tan (\mathrm{x} + \mathrm{C})$, C is an arbitrary

The integrating factor of the first order differential equation

${\mathrm{x}}^2\left({\mathrm{x}}^2 - 1\right)\frac{d\mathrm{y}}{d\mathrm{x}} + \mathrm{x}\left({\mathrm{x}}^2 + 1\right)\mathrm{y} = {\mathrm{x}}^2 - 1$ is

• e^x

• $\mathrm{x} - \frac{1}{\mathrm{x}}$

• $\mathrm{x} + \frac{1}{\mathrm{x}}$

• $\frac{1}{{\mathrm{x}}^2}$

If f(x) = ${\int }_{- 1}^{\mathrm{x}}\left|\mathrm{t}\right|d\mathrm{t}$, then for any $\mathrm{x} \ge 0$, f(x) is equal to

• $\frac{1}{2}\left(1 - {\mathrm{x}}^2\right)$

• $\left(1 - {\mathrm{x}}^2\right)$

• $\frac{1}{2}\left(1 + {\mathrm{x}}^2\right)$

• $\left(1 + {\mathrm{x}}^2\right)$

Let I = ${\int }_0^{100\mathrm{\pi }}\sqrt{\left(1 - \cos \left(2\mathrm{x}\right)\right)}d\mathrm{x}$, then

• I = 0

• I = 200$\sqrt{2}$

• I = $\mathrm{\pi }\sqrt{2}$

• I = 100