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Question: If \[\int {\dfrac{{dx}}{{{x^3}{{(1 + {x^6})}^{\dfrac{2}{3}}}}}} = f(x){(1 + {x^{ - 6}})^{\dfrac{1}{3...

If dxx3(1+x6)23=f(x)(1+x6)13+C\int {\dfrac{{dx}}{{{x^3}{{(1 + {x^6})}^{\dfrac{2}{3}}}}}} = f(x){(1 + {x^{ - 6}})^{\dfrac{1}{3}}} + C, C is a constant of integration, then f(x)f(x) is equal to
A. 12 - \dfrac{1}{2}
B. 16 - \dfrac{1}{6}
C. 6x - \dfrac{6}{x}
D. x2 - \dfrac{x}{2}

Explanation

Solution

We have to choose the correct option which is equal to f(x)f(x). They give the relation of f(x)f(x). We find the value of f(x)f(x), at first, we will integrate the given integration.
For integration, we will apply a substitution method.
According to the substitution method, a given integral f(x)dx\int {f(x)} dx can be transformed into another form by changing the independent variable xx to tt.
Then comparing the given problem we can find the value of f(x)f(x).

Complete step-by-step answer:
It is given that; dxx3(1+x6)23=f(x)(1+x6)13+C\int {\dfrac{{dx}}{{{x^3}{{(1 + {x^6})}^{\dfrac{2}{3}}}}}} = f(x){(1 + {x^{ - 6}})^{\dfrac{1}{3}}} + C, C is a constant of integration
We have to find the value of f(x)f(x).
dxx3(1+x6)23\Rightarrow \int {\dfrac{{dx}}{{{x^3}{{(1 + {x^6})}^{\dfrac{2}{3}}}}}}
The given integration can be written as,
dxx3.x4(1+x6)23\Rightarrow \int {\dfrac{{dx}}{{{x^3}.{x^4}{{(1 + {x^{ - 6}})}^{\dfrac{2}{3}}}}}}
We know that, by property of index, am.an=am+n{a^m}.{a^n} = {a^{m + n}}
So, we have,
dxx7(1+x6)23 (1)\Rightarrow \int {\dfrac{{dx}}{{{x^7}{{(1 + {x^{ - 6}})}^{\dfrac{2}{3}}}}}} \ldots {\text{ }}\left( 1 \right)
Now we apply a substitution.
Let us take,
(1+x6)13=t\Rightarrow {(1 + {x^{ - 6}})^{\dfrac{1}{3}}} = t
Differentiate with respect to xx, we get,
13(1+x6)23.(6x7)dx=dt\Rightarrow \dfrac{1}{3}{(1 + {x^{ - 6}})^{\dfrac{{ - 2}}{3}}}.( - 6{x^{ - 7}})dx = dt
Simplifying we get,
(1+x6)23.(2x7)dx=dt\Rightarrow {(1 + {x^{ - 6}})^{\dfrac{{ - 2}}{3}}}.( - 2{x^{ - 7}})dx = dt
Simplifying again we get,
(1+x6)23.x7dx=12dt\Rightarrow {(1 + {x^{ - 6}})^{\dfrac{{ - 2}}{3}}}.{x^{ - 7}}dx = - \dfrac{1}{2}dt
Now, substitute this value in (1) we get,
12dt\Rightarrow \int {\dfrac{{ - 1}}{2}} dt
Integrating we get,
t2+C\Rightarrow \dfrac{{ - t}}{2} + C
Substitute (1+x6)13=t{(1 + {x^{ - 6}})^{\dfrac{1}{3}}} = t in the above equation we get,
12(1+x6)13+C\Rightarrow \dfrac{{ - 1}}{2}{(1 + {x^{ - 6}})^{\dfrac{1}{3}}} + C
Comparing with the given question we get,
f(x)=12f(x) = \dfrac{{ - 1}}{2}
\therefore Hence, the correct answer is 12 - \dfrac{1}{2}.

So, the correct answer is “Option A”.

Note: According to the substitution method, a given integral f(x) dx\int {f(x){\text{ }}dx} can be transformed into another form by changing the independent variable xx to tt.
It is important to note here that you should make the substitution for a function whose derivative also occurs in the integrand.
Integrating constant is added to the function obtained by evaluating the indefinite integral of a given function, indicating that all indefinite integrals of the given function differ by, at most, a constant.