How do you integrate #int 1/(sqrtx(sqrtx+1)^2)# using substitution?

Answer 1

Christopher Allers

The answer is #=-2/(sqrtx+1)+C#

We need

#intx^ndx=x^(n+1)/(n+1)+C (n!=-1)#

We use the substitution

#u=sqrtx+1#

#du=dx/(2sqrtx)#

Therefore,

#intdx/(sqrtx(1+sqrtx)^2)=2int(du)/u^2#

#=2*-1/u#

#=-2/(sqrtx+1)+C#

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Answer 2

William Abdalla

To integrate the given expression using substitution, we start by letting ( u = \sqrt{x} + 1 ). Then, we differentiate ( u ) with respect to ( x ) to find ( du/dx ). After that, we solve for ( dx ) in terms of ( du ) and ( x ). Substituting the expressions for ( u ), ( du/dx ), and ( dx ) into the integral, we rewrite the integral in terms of ( u ). This allows us to simplify the expression and perform the integration with respect to ( u ). Finally, we substitute back the expression for ( u ) in terms of ( x ) to obtain the final result.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.