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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. $\dfrac{1}{2}\log \left| {\sqrt x - 1} \right| + C$

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Last updated date: 19th Jun 2024
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Answer
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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)$

Complete answer:
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.