Question

Solve the integral to find the value of ( α - β ).

Answer 1

Question :

$\sf\int\dfrac{dx}{(1+\sqrt{x})^{2010}}=2[\dfrac{1}{\alpha(1+\sqrt{x})^{\alpha}}-\dfrac{1}{\beta(1+\sqrt{x})^{\beta}}]+c$

$\sf{where\:\alpha,\beta>0}$

Then find value of $\sf\:(\alpha-\beta)$

Solution :

We have,

$\sf\int\dfrac{dx}{(1+\sqrt{x})^{2010}}=2[\dfrac{1}{\alpha(1+\sqrt{x})^{\alpha}}-\dfrac{1}{\beta(1+\sqrt{x})^{\beta}}]+c...(1)$

Now , let's solve the problem by substitution method

$\sf\:\dfrac{dx}{(1+\sqrt{x})^{2010}}$

Let $\sf(1+\sqrt{x})=v...(2)$

Now , Differentiate it with respect to x ,then ;

$\sf\dfrac{dv}{dx}=\dfrac{1}{2\sqrt{x}}$

$\sf\:dx=2\sqrt{x}\:dv$

From (2)

$\sf(1+\sqrt{x})=v$

$\sf\sqrt{x}=v-1$

Thus,

$\sf\int\dfrac{dx}{(1+\sqrt{x})^{2010}}$

$\sf\int\dfrac{1}{(v)^{2010}}\times2\sqrt{x}\:dv$

$\sf\int\dfrac{dx}{(v)^{2010}}\times2(v-1)dv$

$\sf\int2\times\dfrac{v-1}{v^{2010}}dv$

$\sf\:2\int\dfrac{v-1}{v^{2010}}dv$

$\sf\:2[\int\dfrac{v}{v^{2010}}dv-\int\dfrac{1}{v^{2010}}dv]$

$\sf\:2[\int\:v^{(1-2010)}dv-\int\:v^{(-2010)}dv]$

$\sf\:2[\int\:v^{-2009}dv-\int\:v^{-2010}dv]$

$\sf\pink{We\:Know\:that}$

$\rm\int\:x^n\:dx=\dfrac{x{}^{n+1}}{n+1}+c$ Then ,

$\sf\:2[\int\:v^{-2009}dv-\int\:v^{-2010}dv]$

$\sf=2[\dfrac{v^{-2009+1}}{(-2009+1)}-\dfrac{v^{-2010+1}}{(-2010+1)}]+c$

$\sf=2[\dfrac{v^{-2008}}{-2008}-\dfrac{v^{-2009}}{(-2009)}]+c$

$\sf=2[\dfrac{1}{-2008v^{2008}}-\dfrac{1}{(-2009)v^{2009}}]+c$

On Rearranging the terms

$\sf=2[\dfrac{1}{2009v^{2009}}-\dfrac{1}{2008v^{2008}}]+c$

$\sf=2[\dfrac{1}{2009(1+\sqrt{x})^{2009}}-\dfrac{1}{2008(1+\sqrt{x})^{2008}}]+c...(3)$

$\sf\purple{Now\:,Compare\:(1)\:and\:(3)}$

Then ,

$\sf2[\dfrac{1}{2009(1+\sqrt{x})^{2009}}-\dfrac{1}{2008(1+\sqrt{x})^{2008}}]+c=2[\dfrac{1}{\alpha(1+\sqrt{x})^{\alpha}}-\dfrac{1}{\beta(1+\sqrt{x})^{\beta}}]+c$

Thus,

$\sf\alpha=2009$

and $\sf\beta=2008$

Then ,$\sf\alpha-\beta=2009-2008$

$\sf\blue{\alpha-\beta=1}$

_____________________

$\sf\green{More\:About\:the\:topic}$

Integration by substitution :

The given integral $\sf\int\:f(x)dx$ can be transformed into another by changing the independent variable x to t by substituting x = g(t)