Question

\[\int {\dfrac{1}{{{x^2}{{({x^4} + 1)}^{3/4}}}}} dx\] equals to
A. \[{\left( {1 + \dfrac{1}{{{x^4}}}} \right)^{1/4}} + c\]
B. \[{\left( {{x^4} + 1} \right)^{1/4}} + c\]
C. \[{\left( {1 - \dfrac{1}{{{x^4}}}} \right)^{1/4}} + c\]
D. \[ - {\left( {1 + \dfrac{1}{{{x^4}}}} \right)^{1/4}} + c\]

Answer 1

Hint:We solve this question by making the denominator into simpler form by taking common \[{x^4}\] inside the bracket and then substitute the whole quantity inside the bracket as a single variable to make the integration easier.
* Formula for differentiation is \[\dfrac{{d({x^m})}}{{dx}} = m{x^{m - 1}}\]
* Formula for integration is \[\int {{x^m}dx = \dfrac{{{x^{m + 1}}}}{m}} + c\]

Complete step-by-step answer:
We assume the integral equal to \[I\].
\[I = \int {\dfrac{1}{{{x^2}{{({x^4} + 1)}^{3/4}}}}} dx\] \[...(i)\]
First we solve \[{({x^4} + 1)^{3/4}}\] by breaking it.
Since, we know the property of exponents \[{a^{mn}} = [{({a^m})^n}]\]
We can write \[\dfrac{3}{4} = \dfrac{1}{4} \times 3\]
Therefore, \[{({x^4} + 1)^{3/4}} = {\left[ {{{({x^4} + 1)}^{\dfrac{1}{4}}}} \right]^3}\] \[...(ii)\]
Taking \[{x^4}\] common from the bracket \[{x^4} + 1\], we can write it as \[{x^4}(1 + \dfrac{1}{{{x^4}}})\]
Substitute the value \[{x^4} + 1 = {x^4}(1 + \dfrac{1}{{{x^4}}})\] in the equation \[(ii)\]
\[{({x^4} + 1)^{3/4}} = {\left[ {{{\left( {{x^4}(1 + \dfrac{1}{{{x^4}}})} \right)}^{\dfrac{1}{4}}}} \right]^3}\]
Using the property, \[{(ab)^m} = {a^m} \times {b^m}\] , take the power \[\dfrac{1}{4}\] on both the elements inside the bracket.
                   \[ = {\left[ {{{({x^4})}^{1/4}} \times {{(1 + \dfrac{1}{{{x^4}}})}^{1/4}}} \right]^3}\]
Using the property \[{a^{mn}} = [{({a^m})^n}]\]solve the powers.
                   \[
   = {\left[ {{x^{4 \times \dfrac{1}{4}}} \times {{(1 + \dfrac{1}{{{x^4}}})}^{1/4}}} \right]^3} \\
   = {\left[ {x \times {{(1 + \dfrac{1}{{{x^4}}})}^{1/4}}} \right]^3} \\
 \]
Using the property, \[{(ab)^m} = {a^m} \times {b^m}\] , take the power \[3\] on both the elements inside the bracket.
                  \[ = {x^3}{\left[ {{{(1 + \dfrac{1}{{{x^4}}})}^{1/4}}} \right]^3}\]
Using the property \[{a^{mn}} = [{({a^m})^n}]\]solve the powers.
                  \[
   = {x^3}{\left[ {1 + \dfrac{1}{{{x^4}}}} \right]^{3 \times \dfrac{1}{4}}} \\
   = {x^3}{\left[ {1 + \dfrac{1}{{{x^4}}}} \right]^{\dfrac{3}{4}}} \\
 \]
So, \[{({x^4} + 1)^{3/4}} = {x^3}{\left[ {1 + \dfrac{1}{{{x^4}}}} \right]^{\dfrac{3}{4}}}\]
Substitute the value of \[{({x^4} + 1)^{3/4}}\]in equation \[(i)\]
\[I = \int {\dfrac{1}{{{x^2} \times {x^3}{{\left( {1 + \dfrac{1}{{{x^4}}}} \right)}^{\dfrac{3}{4}}}}}} dx\]
    \[ = \int {\dfrac{1}{{{x^5}{{\left( {1 + \dfrac{1}{{{x^4}}}} \right)}^{\dfrac{3}{4}}}}}} dx\] \[...(iii)\]
Now, let us assume \[\left( {1 + \dfrac{1}{{{x^4}}}} \right) = t\]
Then by differentiating both the sides, we get
\[
  0 + ( - 4)({x^{ - 4 - 1}})dx = dt \\
   - 4{x^{ - 5}}dx = dt \\
 \]
Shifting the value \[ - 4\] to Denominator of RHS
\[{x^{ - 5}}dx = \dfrac{{dt}}{{( - 4)}}\]
Now, we can write \[{x^{ - n}} = \dfrac{1}{{{x^n}}}\]
\[\dfrac{{dx}}{{{x^5}}} = \dfrac{{dt}}{{( - 4)}}\] \[...(iv)\]
Substitute the values in the equation \[(iii)\]
\[I = \int {\dfrac{1}{{{{\left( t \right)}^{\dfrac{3}{4}}}}}} .\dfrac{{dt}}{{( - 4)}}\]
Since we can bring out the constant in the denominator, the integral becomes
\[I = \dfrac{1}{{ - 4}}\int {\dfrac{1}{{{{\left( t \right)}^{\dfrac{3}{4}}}}}} dt\]
Now, we can write \[{x^{ - n}} = \dfrac{1}{{{x^n}}}\]
\[I = - \dfrac{1}{4}\int {{t^{ - \dfrac{3}{4}}}} dt\]
Now we solve the integration by the method \[\int {{x^m}dx = \dfrac{{{x^{m + 1}}}}{m}} + c\]
\[I = - \dfrac{1}{4}\left[ {\dfrac{{{t^{\dfrac{{ - 3}}{4} + 1}}}}{{^{\dfrac{{ - 3}}{4} + 1}}}} \right] + c\]
Take LCM of pwers in the bracket .
\[
  I = - \dfrac{1}{4}\left[ {\dfrac{{{t^{\dfrac{{ - 3 + 4}}{4}}}}}{{^{\dfrac{{ - 3 + 4}}{4}}}}} \right] + c \\
  I = - \dfrac{1}{4}\left[ {\dfrac{{{t^{\dfrac{1}{4}}}}}{{^{\dfrac{1}{4}}}}} \right] + c \\
 \]

 Multiply both numerator and denominator by \[4\].
\[
  I = - \dfrac{1}{4} \times 4\left[ {\dfrac{{{t^{\dfrac{1}{4}}}}}{{^{\dfrac{{4 \times 1}}{4}}}}} \right] + c \\
  I = - \dfrac{1}{4} \times 4\left[ { \times {t^{\dfrac{1}{4}}}} \right] + c \\
  I = - {t^{\dfrac{1}{4}}} + c \\
 \]
Now put back the value we assumed, \[\left( {1 + \dfrac{1}{{{x^4}}}} \right) = t\] .
\[I = - {\left( {1 + \dfrac{1}{{{x^4}}}} \right)^{\dfrac{1}{4}}} + c\]
Thus, option A is correct.

Note:Students are likely to get confused in taking the values common out of the bracket, they should try to do one step at a time to avoid mistakes. We can always use the properties of exponents to break and group together the powers as required by the question. Also, students should know which part to assume as a new variable to make the integration easy.