Challenging Question:-
Solve the problem:
[tex]∫ \frac{1}{ \sqrt{x} (x+1)}dx[/tex]
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2 thoughts on “Challenging Question:- <br />Solve the problem:<br />[tex]∫ \frac{1}{ \sqrt{x} (x+1)}dx[/tex]<br />​”

1. Answer:–

   The correct answer of your question is 2 arc tan ( rootx +c)

   Explanation:–

   ∫f(x)dx=F(x)+C,ifF′(x)=f(x). In this definition, the ∫ is called the integral symbol, f(x) is called the integrand, x is called the variable of integration, dx is called the differential of the variable x, and C is called the constant of integration.

   However, using substitution to evaluate a definite integral requires a change to the limits of integration. If we change variables in the integrand, the limits of integration change as well. Let u=g(x) and let g′ be continuous over an interval [a,b], and let f be continuous over the range of u=g(x).

   The fundamental use of integration is as a continuous version of summing. The extra C, called the constant of integration, is really necessary, since after all differentiation kills off constants, which is why integration and differentiation are not exactly inverse operations of each other.

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2. Answer:

   [tex]\\\tt 2 \tan^-1 \sqrt x +C[/tex]

   Step-by-step explanation:

   Let x = [tex]\\\tt t^2[/tex]

   Now dx = 2t dt

   Substitute this to get,

   [tex]\\\tt \int \dfrac{1}{t(t^2 +1)} .2t dt \\\\\tt \implies \int \dfrac{2}{(t^2 +1)} dt\\\\\tt \implies 2 \tan^{-1} t +C \\\\\tt \implies 2 \tan^{-1} \sqrt x +C[/tex]

   Relation used:

   [tex]\\\tt \int \dfrac{1}{1+x^2} = \tan^{-1}x +C[/tex]

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