Question

The value of the integral $ \int {\left[ {\dfrac{{1 + {x^4}}}{{{{\left( {1 - {x^4}} \right)}^{\dfrac{3}{2}}}}}} \right]dx} = $
1) $ \dfrac{x}{{\sqrt {1 - {x^4}} }} + C $
2) $ \dfrac{{ - x}}{{\sqrt {1 - {x^4}} }} + C $
3) $ \dfrac{{2x}}{{\sqrt {1 - {x^4}} }} + C $
4) $ \dfrac{{ - 2x}}{{\sqrt {1 - {x^4}} }} + C $

Answer 1

Hint: We will first have to operate the given function accordingly to change it to a form such that we can easily integrate it. Then we would have to assume a suitable function as a variable such as $ t $ , and change the function accordingly and perform suitable operations to find the integration to find the result. Then finally we would replace the original variable to get the answer back in the terms of the original variable.

Complete step-by-step answer:
Given, $ \int {\left[ {\dfrac{{1 + {x^4}}}{{{{\left( {1 - {x^4}} \right)}^{\dfrac{3}{2}}}}}} \right]dx} $
Let $ I = \int {\left[ {\dfrac{{1 + {x^4}}}{{{{\left( {1 - {x^4}} \right)}^{\dfrac{3}{2}}}}}} \right]dx} $
Now, dividing both numerator and denominator by $ {x^3} $ , we get,
 $ \Rightarrow I = \int {\left[ {\dfrac{{\dfrac{{1 + {x^4}}}{{{x^3}}}}}{{\dfrac{{{{\left( {1 - {x^4}} \right)}^{\dfrac{3}{2}}}}}{{{x^3}}}}}} \right]dx} $
Therefore, further simplifying, we get,
 $ \Rightarrow I = \int {\left[ {\dfrac{{\dfrac{1}{{{x^3}}} + x}}{{{{\left( {\dfrac{1}{{{x^2}}} - {x^2}} \right)}^{\dfrac{3}{2}}}}}} \right]dx} $
Now, let us assume, $ \dfrac{1}{{{x^2}}} - {x^2} = t $
Now, differentiating both sides, we get,
 $ \Rightarrow \left( {\dfrac{{ - 2}}{{{x^3}}} - 2x} \right)dx = dt $
Taking $ - 2 $ common from the left hand side, we get,
 $ \Rightarrow - 2\left( {\dfrac{1}{{{x^3}}} + x} \right)dx = dt $
Now, dividing both sides by $ - 2 $ , we get,
 $ \Rightarrow \left( {\dfrac{1}{{{x^3}}} + x} \right)dx = \dfrac{{ - 1}}{2}dt $
Now, substituting these values in $ I $ , we get,
 $ \Rightarrow I = \int {\left[ {\dfrac{1}{{{t^{\dfrac{3}{2}}}}}} \right]\left( {\dfrac{{ - 1}}{2}} \right)dt} $
 $ \Rightarrow I = - \dfrac{1}{2}\int {\dfrac{{dt}}{{{t^{\dfrac{3}{2}}}}}} $
Now, we know, the power rule of integration $ \int {{x^n}dx} = \left[ {\dfrac{{{x^{n + 1}}}}{{n + 1}}} \right] + C $ .
Using this formula in $ I $ , we get,
 $ \Rightarrow I = - \dfrac{1}{2}\int {{t^{\dfrac{{ - 3}}{2}}}} dt $
 $ \Rightarrow I = - \dfrac{1}{2}\left[ {\dfrac{{{t^{\dfrac{{ - 3}}{2} + 1}}}}{{\dfrac{{ - 3}}{2} + 1}}} \right] + C $
Where, $ C $ is the constant of integration.
Further simplifying, we get,
 $ \Rightarrow I = - \dfrac{1}{2}\left[ {\dfrac{{{t^{\dfrac{{ - 1}}{2}}}}}{{\dfrac{{ - 1}}{2}}}} \right] + C $
Doing the calculations, we get,
 $ \Rightarrow I = {t^{\dfrac{{ - 1}}{2}}} + C $
Now, replacing back $ t $ in terms of $ x $ , we get,
 $ \Rightarrow I = {\left( {\dfrac{1}{{{x^2}}} - {x^2}} \right)^{\dfrac{{ - 1}}{2}}} + C $
Taking LCM of the expression, we get,
 $ \Rightarrow I = {\left( {\dfrac{{1 - {x^4}}}{{{x^2}}}} \right)^{\dfrac{{ - 1}}{2}}} + C $
We can write it as,
 $ \Rightarrow I = {\left( {\dfrac{{{x^2}}}{{1 - {x^4}}}} \right)^{\dfrac{1}{2}}} + C $
Now, taking square root, we get,
 $ \Rightarrow I = \left( {\dfrac{x}{{\sqrt {1 - {x^4}} }}} \right) + C $
Therefore, the required answer we found is $ \int {\left[ {\dfrac{{1 + {x^4}}}{{{{\left( {1 - {x^4}} \right)}^{\dfrac{3}{2}}}}}} \right]dx} = \left( {\dfrac{x}{{\sqrt {1 - {x^4}} }}} \right) + C $
Hence, the correct option is 1.
So, the correct answer is “Option 1”.

Note: The part where we make mistakes the most is in assuming the function in the form of another variable to substitute in the original question. A simpler way to keep in mind what to do is that, we have to convert the numerator and denominator in such a way that the derivative of the denominator is the denominator or vice-versa. There are further many cases, which can be kept in grasp by practice.