Question

The value of $\int{\dfrac{dx}{x\sqrt{1-{{x}^{3}}}}}$ is equals to
(a) $\dfrac{1}{3}\ln \left| \dfrac{\sqrt{1-{{x}^{3}}}-1}{\sqrt{1-{{x}^{3}}}+1} \right|+C$
(b)$\dfrac{1}{3}\ln \left| \dfrac{\sqrt{1-{{x}^{3}}}+1}{\sqrt{1-{{x}^{3}}}-1} \right|+C$
(c)$\dfrac{1}{3}\ln \left| \dfrac{1}{\sqrt{1-{{x}^{3}}}} \right|+C$
(d)$\dfrac{1}{3}\ln \left| 1-{{x}^{3}} \right|+C$

Answer 1

Hint: Now, to solve this question what we will do is, we will let $\sqrt{1-{{x}^{3}}}=t$, and on differentiating on both sides and on rearranging, we will substitute values in integral I, then by using some standard integration formula, we will end up with final answer.

Complete step by step answer:
Let, $I=\int{\dfrac{dx}{x\sqrt{1-{{x}^{3}}}}}$
Now, let $\sqrt{1-{{x}^{3}}}=t$
Differentiating both side by using formula of differentiation such as $\dfrac{d}{dx}\sqrt{x}=\dfrac{1}{2\sqrt{x}}$ and $\dfrac{d}{dx}({{x}^{n}})=n{{x}^{n-1}}$ also, chain rule of differentiation says that if y = f(g(x)), then $\dfrac{d}{dx}(f(g(x))=f'(g(x))\cdot g'(x)$ , we get
$\dfrac{-3{{x}^{2}}}{2\sqrt{1-{{x}^{3}}}}dx=dt$ ,
Or, re – writing the terms we get
$\dfrac{1}{\sqrt{1-{{x}^{3}}}}dx=-\dfrac{2}{3{{x}^{2}}}dt$
So, substituting $\dfrac{1}{\sqrt{1-{{x}^{3}}}}dx=-\dfrac{2}{3{{x}^{2}}}dt$ in I, we get
$I=\int{-\dfrac{2}{3}\dfrac{dt}{x\cdot {{x}^{2}}}}$
Also, on rearranging terms of $\sqrt{1-{{x}^{3}}}=t$, we get${{x}^{3}}=1-{{t}^{2}}$,
Substitituting, ${{x}^{3}}=1-{{t}^{2}}$ in $I=\int{-\dfrac{2}{3}\dfrac{dt}{{{x}^{3}}}}$, we get
$I=\int{-\dfrac{2}{3}\dfrac{dt}{(1-{{t}^{2}})}}$
Or, $I=\dfrac{2}{3}\int{\dfrac{dt}{({{t}^{2}}-1)}}$
Using, algebraic identity ${{a}^{2}}-{{b}^{2}}=(a+b)(a-b)$ , we can write ${{t}^{2}}-1=(t+1)(t-1)$
So, $I=\dfrac{2}{3}\int{\dfrac{dt}{(t+1)(t-1)}}$
Now, we can write $\dfrac{dt}{(t+1)(t-1)}$ as $\dfrac{dt}{(t+1)(t-1)}=\dfrac{1}{2}\left( \dfrac{1}{t-1}-\dfrac{1}{t+1} \right)dt$
Putting $\dfrac{dt}{(t+1)(t-1)}=\dfrac{1}{2}\left( \dfrac{1}{t-1}-\dfrac{1}{t+1} \right)dt$ in $I=\dfrac{2}{3}\int{\dfrac{dt}{(t+1)(t-1)}}$, we get
$I=\dfrac{2}{3}.\dfrac{1}{2}\int{\left( \dfrac{1}{t-1}-\dfrac{1}{t+1} \right)dt}$
On simplifying, we get
$I=\dfrac{1}{3}\int{\left( \dfrac{1}{t-1}-\dfrac{1}{t+1} \right)dt}$
Or, $I=\dfrac{1}{3}\left( \int{\dfrac{1}{t-1}dt-\int{\dfrac{1}{t+1}d}t} \right)$
We know that, $\int{\dfrac{1}{x+a}dx=\ln (x+a)+C}$ , where a is any real number,
So, \[\int{\dfrac{1}{t-1}dt}\] will be equals to \[\int{\dfrac{1}{t-1}dt}=\ln (t-1)+C\] and \[\int{\dfrac{1}{t+1}dt}\] will be equals to \[\int{\dfrac{1}{t+1}dt}=\ln (t+1)+C\].
Putting, values of \[\int{\dfrac{1}{t-1}dt}=\ln (t-1)+C\] and \[\int{\dfrac{1}{t+1}dt}=\ln (t+1)+C\] in $I=\dfrac{1}{3}\left( \int{\dfrac{1}{t-1}dt-\int{\dfrac{1}{t+1}d}t} \right)$, we get
$I=\dfrac{1}{3}\left( \ln (t-1)+C-\ln (t+1)+C \right)$
Now, we know that ${{\ln }_{a}}(M)-{{\ln }_{a}}(N)={{\ln }_{a}}\left( \dfrac{M}{N} \right)$
So, $\ln (t-1)-\ln (t+1)=\ln \left( \dfrac{t-1}{t+1} \right)$,
Putting, $\ln (t-1)-\ln (t+1)=\ln \left( \dfrac{t-1}{t+1} \right)$ in $I=\dfrac{1}{3}\left( \log (t-1)+C-\log (t+1)+C \right)$, we get
$I=\dfrac{1}{3}\ln \left| \dfrac{t-1}{t+1} \right|$
As, we assumed that, $\sqrt{1-{{x}^{3}}}=t$
So, substituting, value of $\sqrt{1-{{x}^{3}}}=t$, back in integral $I=\dfrac{1}{3}\ln \left| \dfrac{t-1}{t+1} \right|$, we get
$I=\dfrac{1}{3}\ln \left| \dfrac{\sqrt{1-{{x}^{3}}}-1}{\sqrt{1-{{x}^{3}}}+1} \right|+C$

So, the correct answer is “Option A”.

Note: Always remember some of the standard formula of indefinite integrals such as$\int{xdx=\dfrac{{{x}^{2}}}{2}+C}$, $\int{\dfrac{dx}{{{x}^{2}}-{{a}^{2}}}=\dfrac{1}{2a}\ln \left| \dfrac{x-a}{x+a} \right|}+C$ and $\int{\dfrac{1}{x+a}dx=\ln (x+a)+C}$ also, basics of differentiation such as $\dfrac{d}{dx}\sqrt{x}=\dfrac{1}{2\sqrt{x}}$, $\dfrac{d}{dx}({{x}^{n}})=n{{x}^{n-1}}$ and chain rule of differentiation which states that y = f(g(x)), then $\dfrac{d}{dx}(f(g(x))=f'(g(x))\cdot g'(x)$. Do not forget to add constants in the final answer as we are solving indefinite integral not definite integral. Try not to make calculation errors.