Question

$\displaystyle\sf x = 3+2\sqrt{2}$
$\displaystyle\sf \sqrt{x}-\dfrac{1}{\sqrt{x}}$??
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Answer 1

Answer:

Force is an external agent capable of changing the state of rest or motion of a particular body. It has a magnitude and a direction. The direction towards which the force is applied is known as the direction of the force and the application of force is the point where force is applied.

The Force can be measured using a spring balance. The SI unit of force is Newton(N).

Common symbols: F→, F

SI unit: Newton

In SI base units: kg·m/s2

Other units: dyne, poundal, pound-force, kip, kilo pond

Derivations from other quantities:  F = m a

Dimension: LMT-2

Explanation:

Answer 2

ANSWER :-

$\displaystyle \int \dfrac{\sqrt{x}}{1+x}$

We will solve the above problem of integration by

using substitution method :-

$⟹let \: \sqrt{x} = tlet$

$⟹(x = {t}^{2} )(x=t$

$⟹{x}^{ \frac{1}{2} } = tx$

$⟹\dfrac{1}{2} {x}^{ \frac{ - 1}{2} } dx = dt$

$⟹\dfrac{dx}{2 \sqrt{x} } = dt$

$⟹dx = 2 \sqrt{x} dtdx=2$

$⟹dx = 2 t \: dtdx=2tdt$

Now , we get :-

$⟹\displaystyle \int \dfrac{2 {t}^{2} \: dt}{1 + {t}^{2} }$

$⟹2\displaystyle \int \dfrac{ {t}^{2} + 1 - 1\: }{1 + {t}^{2} }\:dt2∫$

$⟹2\displaystyle \int \dfrac{ {t}^{2} + 1 \: }{1 + {t}^{2} }\:dt - 2\displaystyle \int \dfrac{ 1\: }{1 + {t}^{2} }\:dt2∫$

$⟹2\displaystyle \int1 \:dt - 2\displaystyle \int \dfrac{ 1\: }{1 + {t}^{2} }\:dt2∫1dt−2∫ </p><p>$

$⟹2t - 2 {\tan}^{ - 1} (t) + c2t−2tan$

where c = constant

Now put t = √x

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

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Learn more :-

∫ 1 dx = x + C

∫ sin x dx = – cos x + C

∫ cos x dx = sin x + C

∫ sec2 dx = tan x + C

∫ csc2 dx = -cot x + C

∫ sec x (tan x) dx = sec x + C

∫ csc x ( cot x) dx = – csc x + C

∫ (1/x) dx = ln |x| + C

∫ ex dx = ex+ C

∫ ax dx = (ax/ln a) + C

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