∫$$dx(1+x)2010=21$$α(1+x)α−$$1$$β(1+x)β$$+C$$ where α,β>$$0$$ then α$$-$$β is

∫$$xx(1+x)2010dx=2$$∫t−$$1t2010dt($$ put $$t=1+x)=212009t2009$$−$$12008t2008+C$$α$$=2009$$β$$=2008$$α−β$$=1$$

Methods of integration

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Question

$\int \frac{d x}{(1 + \sqrt{x})^{2010}} = 2 [\frac{1}{α (1 + \sqrt{x})^{α}} - \frac{1}{β (1 + \sqrt{x})^{β}}] + C where α, β > 0 then α-β is$

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Solution

$\begin{array}{l} \int \frac{\sqrt{x}}{\sqrt{x} (1 + \sqrt{x})^{2010}} d x = 2 \int \frac{t - 1}{t^{2010}} d t (put t = 1 + \sqrt{x}) \\ = 2 [\frac{1}{2009 t^{2009}} - \frac{1}{2008 t^{2008}}] + C \\ α = 2009 β = 2008 \\ α - β = 1 \end{array}$

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