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Integrals

Multiple Choice Questions

971.

$If \int \frac{dx}{xlog (x - 2) \log (x - 3)} = I + C, then I = ?$

$\frac{1}{x} \log |\frac{\log (x - 3)}{\log (x - 2)}|$
$\log |\frac{\log (x - 3)}{\log (x - 2)}|$
$\log |\frac{\log (x - 2)}{\log (x - 3)}|$
$\log |\log (x - 3) \log (x - 2)|$

972.

$If \int_{0}^{b} \frac{dx}{1 + x^{2}} = \int_{b}^{\infty} \frac{dx}{1 + x^{2}}, then b = ?$

$\tan^{- 1} (\frac{1}{3})$
$\frac{\sqrt{3}}{2}$
$\sqrt{2}$
1

973.

The approximate value of $\int_{1}^{3} \frac{dx}{2 + 3 x}$ using Simpson's rule and dividing the interval [1, 3] into two equal parts is

$\frac{1}{3} \log (\frac{11}{5})$
$\frac{107}{110}$
$\frac{22}{110}$
$\frac{119}{440}$

974.

$If \int \frac{dx}{\sqrt{\sin^{3} (x) \cos (x)}} = g (x) + c, then g (x) = ?$

$- \frac{2}{\sqrt{\cot (x)}}$
$- \frac{2}{\sqrt{\tan (x)}}$
$\frac{2}{\sqrt{\cot (x)}}$
$\frac{2}{\sqrt{\tan (x)}}$

975.
$If \int \frac{dx}{(1 + \sqrt{x}) \sqrt{x - x^{2}}} = \frac{A \sqrt{x}}{\sqrt{1 - x}} + \frac{B}{\sqrt{1 - x}} + C, where C is real constant, then A + B = ?$

3

0

1

2

$Given, \int \frac{dx}{(1 + \sqrt{x}) \sqrt{x - x^{2}}} = \frac{A \sqrt{x}}{\sqrt{1 - x}} + \frac{B}{\sqrt{1 - x}} + C On differentiating both sides, we get \frac{1}{(1 + \sqrt{x}) (x - \sqrt{x})} = \frac{\sqrt{1 - x} (A \frac{1}{2 \sqrt{x}}) - (A \sqrt{x} + B) \frac{1}{2 \sqrt{2 - x}} (- 1)}{{(\sqrt{1 - x})}^{2}} \Rightarrow \frac{1}{(1 + \sqrt{x}) (\sqrt{x - x^{2}})} = \frac{(1 - x) A + Ax + B \sqrt{x}}{(1 - x) 2 \sqrt{x} (\sqrt{1 - x})} \Rightarrow \frac{2 (1 - x)}{(1 + \sqrt{x})} = A + B \sqrt{x} \Rightarrow 2 - 2 \sqrt{x} = A = B \sqrt{x} \Rightarrow A = 2, B = 2 ∴ A - B = 0$

976.

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)

977.

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

$\frac{1}{2}$
3
2
1

978.

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