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Integrals

Multiple Choice Questions

691.

For any integer n $\geq$ 2, let I_n = $\int$ tanⁿ(x)dx. If $I_{n} = \frac{1}{a} \tan^{n - 1} (x) - {bI}_{n - 2} for n \geq 2,$ then the ordered par (a, b) equals to

$(n - 1, \frac{n - 1}{n - 2})$
$(n - 1, \frac{n - 2}{n - 1})$
(n, 1)
(n - 1, 1)

692.
$If \frac{(x^{2} - 1)}{{(x + 1)}^{2} \sqrt{x (x^{2} + x + 1)}} dx = {Atan}^{- 1} (\sqrt{\frac{x^{2} + x + 1}{x}}) C, in which C is a constant, then A = ?$

$\frac{1}{2}$

3

2

1

$Let I = \int \frac{(x^{2} - 1)}{{(x + 1)}^{2} \sqrt{x (x^{2} + x + 1)}} dx Now, \frac{d}{dx} \tan^{- 1} (\sqrt{\frac{(x^{2} + x + 1)}{x}}) = \frac{1}{1 + \frac{(x^{2} + x + 1)}{x}} \times \frac{d}{dx} (\sqrt{\frac{(x^{2} + x + 1)}{x}})$

$= \frac{x}{x^{2} + 2 x + 1} [\frac{\sqrt{x} \times \frac{2 x + 1}{2 \sqrt{(x^{2} + x + 1)}} - \sqrt{(x^{2} + x + 1)} \frac{1}{2 \sqrt{x}}}{{(\sqrt{x})}^{2}}] = \frac{x}{x^{2} + 2 x + 1} [\frac{\sqrt{x} (2 x + 1)}{2 \sqrt{(x^{2} + x + 1)}} - \frac{\sqrt{(x^{2} + x + 1)}}{2 \sqrt{x}}] = \frac{x}{(x^{2} + 2 x + 1)} [\frac{(2 x + 1) x - (x^{2} + x + 1)}{\frac{2 \sqrt{x (x^{2} + x + 1)}}{x}}]$

$= \frac{x}{{(x + 1)}^{2}} [\frac{x^{2} - 1}{2 \times \sqrt{x (x^{2} + x + 1)}}] = \frac{x^{2} - 1}{2 {(x + 1)}^{2} \sqrt{(x^{2} + x + 1)}} ∴ I = \int \frac{d}{dx} (2 \tan^{- 1} (\sqrt{\frac{x^{2} + x + 1}{x}})) = 2 \tan^{- 1} (\sqrt{\frac{x^{2} + x + 1}{x}}) + C ∴ A = 2$

693.

$By the defination of the definite integral, the value of \lim_{n \to \infty} (\frac{1^{4}}{1^{5} + n^{5}} + \frac{2^{4}}{2^{5} + n^{5}} + \frac{3^{4}}{3^{5} + n^{6}} + . . . + \frac{n^{4}}{n^{5} + n^{5}}) is$