If I- -1-x2-2-x3-x2-1 dx- then I equals

The correct option is C 0 nbsp;I=∫ _ 1 ^∞ nbsp; x ^ 2 2 / x ^ 3 √( x ^ 2 1 ) nbsp; dx Put Put x= u ⇒ dx=tan u u du nbsp;I=∫ _ 0 ^π ...

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Byju's Answer

Standard XII

Mathematics

Integration Using Substitution

If I= ∫1∞x2...

Question

If $I = \int_{1}^{\infty} \frac{x^{2} - 2}{x^{3} \sqrt{x^{2} - 1}} d x$ , then $I$ equals

- 1

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0

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π / 2

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π - \sqrt{3}

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Solution

The correct option is C $0$
$I = \int_{1}^{\infty} \frac{x^{2} - 2}{x^{3} \sqrt{x^{2} - 1}} d x$
Put Put $x = sec u \Rightarrow d x = tan u sec u d u$
$I = \int_{0}^{\frac{π}{2}} {cos}^{2} u ({sec}^{2} u - 2) d u = \int_{0}^{\frac{π}{2}} (1 - {sin}^{2} u) ({sec}^{2} u - 2) d u$
$= \int_{0}^{\frac{π}{2}} (2 {sin}^{2} u - {tan}^{2} u + {sec}^{2} u - 2) d u = - \int_{0}^{\frac{π}{2}} d u + 2 \int_{0}^{\frac{π}{2}} {sin}^{2} u d u$
$= - \int_{0}^{\frac{π}{2}} d u - \int_{0}^{\frac{π}{2}} cos 2 u d u + \int_{0}^{\frac{π}{2}} d u = - {[\frac{sin 2 u}{2}]}_{0}^{\frac{π}{2}} = 0$

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If I=∫∞1x2−2x3√x2−1dx, then I equals

The correct option is C 0I=∫∞1x2−2x3√x2−1dxPut Put x=secu⇒dx=tanusecuduI=∫π20cos2u(sec2u−2)du=∫π20(1−sin2u)(sec2u−2)du=∫π20(2sin2u−tan2u+sec2u−2)du=−∫π20du+2∫π20sin2udu=−∫π20du−∫π20cos2udu+∫π20du=−[sin2u2]π20=0

If $I = \int_{1}^{\infty} \frac{x^{2} - 2}{x^{3} \sqrt{x^{2} - 1}} d x$ , then $I$ equals