Question

Find the value of the integration \[\int {\left\{ {\dfrac{1}{{\left[ {{x^2}{{\left( {{x^4} + 1} \right)}^{\dfrac{3}{4}}}} \right]}}} \right\}} dx\]
A) \[\left[ { - \dfrac{{{{\left( {{x^4} + 1} \right)}^{\dfrac{3}{4}}}}}{x}} \right] + c\]
B) \[\left[ { - \dfrac{{{{\left( {1 - {x^4}} \right)}^{\dfrac{1}{4}}}}}{{2x}}} \right] + c\]
C) \[\left[ { - \dfrac{{{{\left( {1 + {x^4}} \right)}^{\dfrac{1}{4}}}}}{x}} \right] + c\]
D) \[\left[ { - \dfrac{{{{\left( {1 + {x^4}} \right)}^{\dfrac{1}{4}}}}}{{{x^2}}}} \right] + c\]
E) \[\left[ { - \dfrac{{{{\left( {1 + {x^4}} \right)}^{\dfrac{1}{2}}}}}{x}} \right] + c\]

Answer 1

Hint: To solve this question we need to follow the substitution process . We have to substitute a unit such that after the substitution integration becomes simple and we can apply the basic rule of integration easily .
FORMULA USED :
\[\dfrac{{d\left( {{x^n}} \right)}}{{dx}} = n{x^{n - 1}}\]
\[\int {{x^n}} = \dfrac{{{x^{n + 1}}}}{{n + 1}} + c\]

Complete step-by-step answer:
We have to substitute a portion of the integrand so that the integration becomes simple . To know what part of the integration to be substituted here for this problem we simplify the integrand first .
We must take\[{x^4}\]common from \[{\left( {{x^4} + 1} \right)^{\dfrac{3}{4}}}\]portion . So denominator of the integrand become \[{x^2} \times {x^3} \times {\left( {1 + \dfrac{1}{{{x^4}}}} \right)^{\dfrac{3}{4}}}\]\[ = {x^5}{\left( {1 + \dfrac{1}{{{x^4}}}} \right)^{\dfrac{3}{4}}}\]
So the integrand becomes \[\int {\left\{ {\dfrac{1}{{\left[ {{x^5}{{\left( {1 + \dfrac{1}{{{x^4}}}} \right)}^{\dfrac{3}{4}}}} \right]}}} \right\}} dx\]
Now we can decide which part we will substitute so that integration becomes simpler.
We will substitute \[\left( {1 + \dfrac{1}{{{x^4}}}} \right)\] as \[u\]and then take derivative of \[\left( {1 + \dfrac{1}{{{x^4}}}} \right)\].
Let us assume \[\left( {1 + \dfrac{1}{{{x^4}}}} \right)\]\[ = u\]
After differentiating both side we get
\[ - \dfrac{4}{{{x^5}}}dx = du\]
\[ \Rightarrow dx = - \dfrac{{{x^5}}}{4}du\]
Putting this value in \[\int {\left\{ {\dfrac{1}{{\left[ {{x^5}{{\left( {1 + \dfrac{1}{{{x^4}}}} \right)}^{\dfrac{3}{4}}}} \right]}}} \right\}} dx\]we get
\[\int {\left\{ {\dfrac{1}{{\left[ {{x^5}{{\left( {1 + \dfrac{1}{{{x^4}}}} \right)}^{\dfrac{3}{4}}}} \right]}}} \right\}} \times - \dfrac{{{x^5}}}{4}du\]
After substitution of \[\left( {1 + \dfrac{1}{{{x^4}}}} \right)\]\[ = u\]
And after cancelling \[{x^5}\]from denominator and numerator we get the integration as \[\int { - \left\{ {\dfrac{1}{{4\left[ {{u^{\dfrac{3}{4}}}} \right]}}} \right\}} du\]
\[ = \int {\dfrac{{ - {u^{ - \dfrac{3}{4}}}}}{4}} du\]
We can now see that the integration became so simple ,easy and less complicated.
Now applying basic integration rule we can solve the integration
\[\int {\dfrac{{ - {u^{ - \dfrac{3}{4}}}}}{4}} du\]
\[
   = - \dfrac{1}{4} \times \dfrac{{{u^{ - \dfrac{3}{4} + 1}}}}{{ - \dfrac{3}{4} + 1}} + c \;
 \]
After simplifying we get \[ = - {u^{\dfrac{1}{4}}} + c\]
\[ = - {u^{\dfrac{1}{4}}} + c\]
\[c\]=integration constant
At last we need to convert the answer in terms of \[x\].
So we will put the value of u in \[ - {u^{\dfrac{1}{4}}} + c\]
\[\left( {1 + \dfrac{1}{{{x^4}}}} \right)\]\[ = u\]
So answer is \[ - {\left( {1 + \dfrac{1}{{{x^4}}}} \right)^{\dfrac{1}{4}}} + c\]
But this answer does not look like any answer in the options . So we will simplify further to get the desired option .
After simplifying we get
\[ = - \dfrac{{{{\left( {{x^4} + 1} \right)}^{\dfrac{1}{4}}}}}{{{{\left( {{x^4}} \right)}^{\dfrac{1}{4}}}}} + c\]
\[ = - \dfrac{{{{\left( {{x^4} + 1} \right)}^{\dfrac{1}{4}}}}}{x} + c\]
Now we can see the answer matches option C.
So, the correct answer is “Option C”.

Note: We need to choose the part which has to be substituted carefully such that it simplifies the integration .For that we simplify the integrand first .
After the integration we have to put the value back of the substitution .