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}}}$$$$ = {x^5}{\left( {1 + \dfrac{1}{{{x^4}}}} \right)^{\dfrac{3}{4}}}
So the integrand becomes \int {\left\\{ {\dfrac{1}{{\left[ {{x^5}{{\left( {1 + \dfrac{1}{{{x^4}}}} \right)}^{\dfrac{3}{4}}}} \right]}}} \right\\}} dx
Now we can decide which part we will substitute so that integration becomes simpler.
We will substitute $\left( {1 + \dfrac{1}{{{x^4}}}} \right)$ as $u$and then take derivative of $\left( {1 + \dfrac{1}{{{x^4}}}} \right)$.
Let us assume \left( {1 + \dfrac{1}{{{x^4}}}} \right)$$$$ = u
After differentiating both side we get
$- \dfrac{4}{{{x^5}}}dx = du$
$\Rightarrow dx = - \dfrac{{{x^5}}}{4}du$
Putting this value in \int {\left\\{ {\dfrac{1}{{\left[ {{x^5}{{\left( {1 + \dfrac{1}{{{x^4}}}} \right)}^{\dfrac{3}{4}}}} \right]}}} \right\\}} dxwe get
\int {\left\\{ {\dfrac{1}{{\left[ {{x^5}{{\left( {1 + \dfrac{1}{{{x^4}}}} \right)}^{\dfrac{3}{4}}}} \right]}}} \right\\}} \times - \dfrac{{{x^5}}}{4}du
After substitution of \left( {1 + \dfrac{1}{{{x^4}}}} \right)$$$$ = 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 .