Question

Solve $\int {\left( {\sqrt[3]{x}} \right)\left( {\sqrt[5]{{1 + \sqrt[3]{{{x^4}}}}}} \right)} dx = $
1) ${\left( {1 + {x^{\dfrac{3}{4}}}} \right)^{\dfrac{6}{5}}} + C$
2) ${\left( {1 + {x^{\dfrac{4}{3}}}} \right)^{\dfrac{6}{5}}} + C$
3) $\dfrac{5}{8}{\left( {1 + {x^{\dfrac{4}{3}}}} \right)^{\dfrac{6}{5}}} + C$
4) $\dfrac{1}{6}{\left( {1 + {x^{\dfrac{4}{3}}}} \right)^6} + C$
5) $\dfrac{{15}}{7}{\left( {1 + {x^{\dfrac{4}{3}}}} \right)^{\dfrac{6}{5}}} + C$

Answer 1

Hint: First we have to assume the function of $x$ in such a way that it becomes easier for us to find the integration in terms of any other variable like $t$. Then if it's still difficult to integrate the function, then we may have to assume it again in terms of another variable such as $z$. Then on integrating the function, we will get an answer in terms of $z$. Then according to the assumed variables we will substitute the original variable and get an integration in the form of $x$.

Complete step-by-step answer:
Given, $\int {\left( {\sqrt[3]{x}} \right)\left( {\sqrt[5]{{1 + \sqrt[3]{{{x^4}}}}}} \right)} dx$.
We can write it as,
$ = \int {\left( {\sqrt[3]{x}} \right)\left( {\sqrt[5]{{1 + \sqrt[3]{{{x^3}.x}}}}} \right)} dx$
$ = \int {\left( {\sqrt[3]{x}} \right)\left( {\sqrt[5]{{1 + x\sqrt[3]{x}}}} \right)} dx$
Let us assume, $\sqrt[3]{x} = t$
Taking cube as power on both sides, we get,
$ \Rightarrow x = {t^3}$
Differentiating both sides, we get,
$ \Rightarrow dx = 3{t^2}dt$
Now, substituting these values in the above equation, we get,
$ = \int {\left( t \right)\left( {\sqrt[5]{{1 + {t^3}.t}}} \right)} 3{t^2}dt$
$ = \int {t.3{t^2}\left( {\sqrt[5]{{1 + {t^4}}}} \right)} dt$
$ = \int {3{t^3}\left( {\sqrt[5]{{1 + {t^4}}}} \right)} dt$
Now, let us assume $1 + {t^4} = z$
Therefore, differentiating both sides, we get,
$ \Rightarrow 4{t^3}dt = dz$
$ \Rightarrow {t^3}dt = \dfrac{{dz}}{4}$
Now, substituting the values in the above equation, we get,
$ = \int {\dfrac{3}{4}\left( {\sqrt[5]{z}} \right)} dz$
We can write it as,
$ = \dfrac{3}{4}\int {{z^{\dfrac{1}{5}}}} dz$
We know, the formula of integration, $\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C$.
Therefore, using this formula in the above function, we get,
$ = \dfrac{3}{4}.\dfrac{{{z^{\dfrac{1}{5} + 1}}}}{{\dfrac{1}{5} + 1}} + C$
Where, $C = $constant of integration.
Now, simplifying, we get,
$ = \dfrac{3}{4}.\dfrac{{{z^{\dfrac{6}{5}}}}}{{\dfrac{6}{5}}} + C$
$ = \dfrac{5}{8}.{z^{\dfrac{6}{5}}} + C$
Now, we know, $z = 1 + {t^4}$, as assumed, so, substituting this in the above equation, we get,
$ = \dfrac{5}{8}.{\left( {1 + {t^4}} \right)^{\dfrac{6}{5}}} + C$
Now, we know, $t = \sqrt[3]{x}$, as assumed, we get,
$ = \dfrac{5}{8}.{\left( {1 + {{\left( {\sqrt[3]{x}} \right)}^4}} \right)^{\dfrac{6}{5}}} + C$
$ = \dfrac{5}{8}.{\left( {1 + {{\left( {{x^{\dfrac{1}{3}}}} \right)}^4}} \right)^{\dfrac{6}{5}}} + C$
$ = \dfrac{5}{8}.{\left( {1 + {x^{\dfrac{4}{3}}}} \right)^{\dfrac{6}{5}}} + C$
Therefore, $\int {\left( {\sqrt[3]{x}} \right)\left( {\sqrt[5]{{1 + \sqrt[3]{{{x^4}}}}}} \right)} dx = \dfrac{5}{8}.{\left( {1 + {x^{\dfrac{4}{3}}}} \right)^{\dfrac{6}{5}}} + C$.
Hence, the correct option is 3.
So, the correct answer is “Option 3”.

Note: Assuming the integration as any other variable is one of the easiest and simplest ways to solve integration. But sometimes it gets more complex and difficult doing so. So, we may have to use other methods like integration by parts etc. Take care of the calculations so as to be sure of the final answer.