If -dx1-x2010-2-1-x-2-1-x-C- where C is a constant of integration and -0- then

I=∫2√(x) dx2√(x)(1+√(x))^2010 Put 1+√(x)=t ⇒12√(x)dx=dt I=2∫(t 1)dtt^2010 =2∫(t^ 2009 t^ 2010)dt =2(t^ 2008 2008 t^ 2009 2009)+C =22009(1+√(x))^2009 22008(1+√( ...

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Byju's Answer

Standard XII

Mathematics

Integration by Substitution

If ∫dx1+√x201...

Question

If $\int \frac{d x}{(1 + \sqrt{x})^{2010}} = \frac{2}{α (1 + \sqrt{x})^{α}} - \frac{2}{β (1 + \sqrt{x})^{β}} + C,$ where $C$ is a constant of integration and $α, β > 0,$ then

| α - β | = 1

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\frac{β + 2}{(2010)^{2}} = \frac{1}{α + 1}

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β, α, 2010

(taken in that order) are in arithmetic progression.

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α + 1 = β + 1 = 2010

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Solution

The correct option is C $β, α, 2010$ (taken in that order) are in arithmetic progression.
$I = \int \frac{2 \sqrt{x} d x}{2 \sqrt{x} (1 + \sqrt{x})^{2010}}$
Put $1 + \sqrt{x} = t \Rightarrow \frac{1}{2 \sqrt{x}} d x = d t$
$I = 2 \int \frac{(t - 1) d t}{t^{2010}}$
$= 2 \int (t^{- 2009} - t^{- 2010}) d t$

$= 2 (\frac{t^{- 2008}}{- 2008} - \frac{t^{- 2009}}{- 2009}) + C$
$= \frac{2}{2009 (1 + \sqrt{x})^{2009}} - \frac{2}{2008 (1 + \sqrt{x})^{2008}} + C$
$∴ α = 2009, β = 2008$ .

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If ∫dx(1+√x)2010=2α(1+√x)α−2β(1+√x)β+C, where C is a constant of integration and α,β>0, then

The correct option is C β,α,2010 (taken in that order) are in arithmetic progression.I=∫2√x dx2√x(1+√x)2010 Put 1+√x=t⇒12√xdx=dt I=2∫(t−1)dtt2010 =2∫(t−2009−t−2010)dt =2(t−2008−2008−t−2009−2009)+C =22009(1+√x)2009−22008(1+√x)2008+C ∴α=2009,β=2008.

If $\int \frac{d x}{(1 + \sqrt{x})^{2010}} = \frac{2}{α (1 + \sqrt{x})^{α}} - \frac{2}{β (1 + \sqrt{x})^{β}} + C,$ where $C$ is a constant of integration and $α, β > 0,$ then