Question

Find the integration of $\int{\dfrac{dx}{x-\sqrt{x}}}$
A. $2\log |\sqrt{x}-1|+C$
B. $2\log |\sqrt{x}+1|+C$
C. $\log |\sqrt{x}-1|+C$
D. $\dfrac{1}{2}\log |\sqrt{x}+1|+C$
E. $\dfrac{1}{2}\log |\sqrt{x}-1|+C$

Answer 1

In the given question, we have an expression which we have to integrate. We will begin by first taking $\sqrt{x}$ common in the denominator and then we will make the numerator almost similar to the denominator by adding and subtracting the required. We will then have, $\int{\dfrac{\sqrt{x}-\left( \sqrt{x}-1 \right)}{\sqrt{x}\left( \sqrt{x}-1 \right)}dx}$. We will separate out the components and integrate it individually and then at the end, we will club it together and solve further to get the final answer. Hence, we will have the value of the given expression.

Complete step by step solution:
According to the given question, we are given an expression which we have to integrate and find the value of the given expression.
The expression we have is,
$\int{\dfrac{dx}{x-\sqrt{x}}}$----(1)
We will begin solving the expression by first taking $\sqrt{x}$ common in the denominator. We will get,
$\Rightarrow \int{\dfrac{dx}{\sqrt{x}\left( \sqrt{x}-1 \right)}}$
Next, we will modify the numerator such that it is somewhat similar to the denominator. That is,
$\Rightarrow \int{\dfrac{1.dx}{\sqrt{x}\left( \sqrt{x}-1 \right)}}$
We will add and subtract $\sqrt{x}$ in the numerator. So, we will get,
$\Rightarrow \int{\dfrac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}\left( \sqrt{x}-1 \right)}dx}$ -----(2)
Now, we will take negative sign common in the numerator such the component formed is similar to the one in the denominator, we get,
$\Rightarrow \int{\dfrac{\sqrt{x}-\left( \sqrt{x}-1 \right)}{\sqrt{x}\left( \sqrt{x}-1 \right)}dx}$
Separating out the fraction in the above equation, we will get,
$\Rightarrow \int{\dfrac{\sqrt{x}}{\sqrt{x}\left( \sqrt{x}-1 \right)}dx-\int{\dfrac{\left( \sqrt{x}-1 \right)}{\sqrt{x}\left( \sqrt{x}-1 \right)}}dx}$-----(3)
We will now cancel out the similar terms present in the numerator and the denominator from both the terms, we will get,
$\Rightarrow \int{\dfrac{1}{\left( \sqrt{x}-1 \right)}dx-\int{\dfrac{1}{\sqrt{x}}}dx}$----(4)
The obtained equation (4) will be integrated separately and then we will merge them and solve further.
We will first solve the $\int{\dfrac{1}{\left( \sqrt{x}-1 \right)}dx}$ part of the equation (4) and we will mark this as equation (5).
Let $t=\sqrt{x}-1$----(5a)
Differentiating both the sides, we get,
$dt=\dfrac{1}{2}{{x}^{\dfrac{1}{2}-1}}dx$
$\Rightarrow dt=\dfrac{1}{2\sqrt{x}}dx$
Writing in terms of ‘dx’, we get,
$\Rightarrow dx=2\sqrt{x}dt$-----(5b)
And we know that, $\sqrt{x}=t+1$----(5c)
Substituting the equations 5a, 5b, 5c in equation (5), we will get,
$\Rightarrow \int{\dfrac{1}{t}2(t+1)dt}$
Taking the constant 2 out, we get,
$\Rightarrow 2\int{\dfrac{1}{t}(t+1)dt}$
$\Rightarrow 2\int{\left( 1+\dfrac{1}{t} \right)dt}$----(5d)
Integrating the equation 5d, we get the expression as,
$\Rightarrow 2\left( t+\log t \right)$
Substituting the value ‘t’ back in the above expression, we get the new expression as,
$\Rightarrow 2\left( \left( \sqrt{x}-1 \right)+\log |\sqrt{x}-1| \right)$-----(5e)
Now, we will solve the second part of the equation (4), which is,
$\int{\dfrac{1}{\sqrt{x}}}dx$
We can write the above expression also as,
$\int{{{x}^{-\dfrac{1}{2}}}}dx$----(6)
We know the formula for integration, which is, $\int{{{x}^{n}}}dx=\dfrac{{{x}^{n+1}}}{n+1}$. Applying this in the equation (6), we get,
$\Rightarrow \dfrac{{{x}^{-\dfrac{1}{2}+1}}}{-\dfrac{1}{2}+1}$----(6a)
Solving the above equation 6a further, we get,
$\Rightarrow \dfrac{{{x}^{\dfrac{1}{2}}}}{\dfrac{1}{2}}$
$\Rightarrow 2\sqrt{x}$----(6b)
We will now merge the two parts of the equation (4). That is, substituting equation (5e) and equation (6b) in the equation (4), we will get,
$\Rightarrow 2\left( \left( \sqrt{x}-1 \right)+\log |\sqrt{x}-1| \right)-2\sqrt{x}$----(7)
Opening up the brackets, we will get,
$\Rightarrow 2\left( \sqrt{x}-1 \right)+2\log |\sqrt{x}-1|-2\sqrt{x}$
Solving the expression further, cancelling out the similar terms, we get the new expression as,
$\Rightarrow 2\sqrt{x}-2+2\log |\sqrt{x}-1|-2\sqrt{x}$
$\Rightarrow 2\log |\sqrt{x}-1|-2$---(8)
The equation (8) is the value of the given expression.
From the given options, our answer matches with option A.
Therefore, the value of the given expression is A. $2\log |\sqrt{x}-1|+C$, where C is a constant.

So, the correct answer is “Option C”.

Note: The integration of the expression can at times be very confusing, so always proceed with the calculations in a clear and step wise form. In equation (4), we did carry out the integration together as it will create a lot of confusion and a slight mistake in integration can prove fatal. So, it is advisable that the integration should be carried out separately for large expression wherever possible.