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# Integration of $\int {\dfrac{{dx}}{{x - \sqrt x }}}$is equal to:A. $2\log \left| {\sqrt x - 1} \right| + C$B. $2\log \left| {\sqrt x + 1} \right| + C$C. $\log \left| {\sqrt x - 1} \right| + C$D. $\dfrac{1}{2}\log \left| {\sqrt x + 1} \right| + C$E. $\dfrac{1}{2}\log \left| {\sqrt x - 1} \right| + C$

Last updated date: 19th Jun 2024
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Hint: according to the question we have to find the integration of $\int {\dfrac{{dx}}{{x - \sqrt x }}}$.
So, first of all we have to let$\sqrt x$= u then we have to differentiate both terms with respect to $x$with the help of the formula that is mentioned below.

Formula used:
$\dfrac{d}{{dx}}\sqrt x = \dfrac{1}{{2\sqrt x }}...........................(A)$
Now, we have to put all the value of $\sqrt x$and $dx$in the given expression$\int {\dfrac{{dx}}{{x - \sqrt x }}}$
After that we have to use the integration formula of $\dfrac{1}{{(a + 1)}}da$that is mentioned below.
$\int {\dfrac{1}{{a + 1}}} da$$= \log \left| {a + 1} \right| + C............................(B)$

Step 1: First of all we have to let the $\sqrt x$=u
Now, differentiate both terms with respect to$x$
$\Rightarrow \dfrac{d}{{dx}}\sqrt x = \dfrac{d}{{dx}}\left( u \right)$
Now, we have to apply the formula (A) that is mentioned in the solution hint.
$\Rightarrow \dfrac{1}{{2\sqrt x }} = \dfrac{{du}}{{dx}} \\ \Rightarrow dx = 2\sqrt x du........................(1) \\$
Step 2: Now, we put the value of $\sqrt x$=u in the expression (1) obtained in step 1
$\Rightarrow dx = 2udu$
Step 3: Now, we put all values obtained in step 1 and step 2 in the given expression $\int {\dfrac{{dx}}{{x - \sqrt x }}}$
$\Rightarrow \int {\dfrac{{2udu}}{{(u + {u^2})}}} \\ \Rightarrow \int {\dfrac{{2udu}}{{u(1 + u)}}} \\ \Rightarrow \int {\dfrac{{2du}}{{(1 + u)}}} \\$
Step 4: Now, we have to apply the formula (B) in the expression mentioned in the step 3.
$\Rightarrow 2\log \left| {u + 1} \right| + C$
Now, put the value of $u$ in terms of $x$.
$\Rightarrow 2\log \left| {\sqrt x + 1} \right| + C$

The integrated values of the given expression $\int {\dfrac{{dx}}{{x - \sqrt x }}}$ $= 2\log \left| {\sqrt x + 1} \right| + C$.

Note:
It is necessary that we have to let the term $\sqrt x$= u and then we have to find the differentiation of the term we let to simplify the expression.
It is necessary that we have let x as ${u^2}$ and then we have to substitute the value in the given expression to find the integration.