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Indefinite Integrals question

2019 · 8 Apr · Shift 2 · Q45
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  5. /2019 · 8 Apr · Shift 2 · Q45

Indefinite Integrals question

2019 · 8 Apr · Shift 2 · Q45

JEE MainMathematicsIndefinite IntegralsMCQ+4 / −1
If ∫dxx3(1+x6)2/3=xf(x)(1+x6)13+C\int {{{dx} \over {{x^3}{{(1 + {x^6})}^{2/3}}}} = xf(x){{(1 + {x^6})}^{{1 \over 3}}} + C}∫x3(1+x6)2/3dx​=xf(x)(1+x6)31​+C where C is a constant of integration, then the function ƒ(x) is equal to
  1. A
    3x2{3 \over {{x^2}}}x23​
  2. B
    −16x3- {1 \over {6{x^3}}}−6x31​
  3. C
    −12x3- {1 \over {2{x^3}}}−2x31​
  4. D
    −12x2- {1 \over {2{x^2}}}−2x21​
View written solutionFree

Correct answer: C

  1. We need to evaluate I=∫dxx3(1+x6)2/3I=\int \frac{dx}{x^3(1+x^6)^{2/3}}I=∫x3(1+x6)2/3dx​ and match it with I=xf(x)(1+x6)1/3+C.I=x f(x)(1+x^6)^{1/3}+C.I=xf(x)(1+x6)1/3+C.

So we try to simplify the integral first.

  1. Substitute u=x−2.u=x^{-2}.u=x−2. Then du=−2x−3 dx⇒x−3dx=−12du.du=-2x^{-3}\,dx \quad \Rightarrow \quad x^{-3}dx=-\frac12 du.du=−2x−3dx⇒x−3dx=−21​du.

Also, 1+x6=1+1u3=u3+1u3.1+x^6=1+\frac{1}{u^3}=\frac{u^3+1}{u^3}.1+x6=1+u31​=u3u3+1​. Hence (1+x6)2/3=(u3+1u3)2/3=(u3+1)2/3u2.(1+x^6)^{2/3}=\left(\frac{u^3+1}{u^3}\right)^{2/3}=\frac{(u^3+1)^{2/3}}{u^2}.(1+x6)2/3=(u3u3+1​)2/3=u2(u3+1)2/3​.

Therefore,

=\int \frac{-\frac12 du}{(u^3+1)^{2/3}/u^2} = -\frac12 \int \frac{u^2\,du}{(u^3+1)^{2/3}}.$$ 3. Now let $$v=u^3+1.$$ Then $$dv=3u^2\,du \quad \Rightarrow \quad u^2\,du=\frac{dv}{3}.$$ So, $$I=-\frac12\int \frac{1}{v^{2/3}}\cdot \frac{dv}{3} =-\frac16\int v^{-2/3}\,dv.$$ 4. Integrate: $$\int v^{-2/3}\,dv=\frac{v^{1/3}}{1/3}=3v^{1/3}.$$ Thus, $$I=-\frac16\cdot 3v^{1/3}+C=-\frac12 v^{1/3}+C.$$ Substituting back, $$I=-\frac12 (u^3+1)^{1/3}+C.$$ Now $u=x^{-2}$, so $$u^3=x^{-6}.$$ Hence, $$I=-\frac12 (1+x^{-6})^{1/3}+C.$$ 5. Rewrite in the required form: $$1+x^{-6}=\frac{1+x^6}{x^6}.$$ Therefore, $$(1+x^{-6})^{1/3}=\left(\frac{1+x^6}{x^6}\right)^{1/3}=\frac{(1+x^6)^{1/3}}{x^2}.$$ So, $$I=-\frac12\cdot \frac{(1+x^6)^{1/3}}{x^2}+C.$$ Now compare with $$I=x f(x)(1+x^6)^{1/3}+C.$$ Thus, $$x f(x)=-\frac{1}{2x^2}$$ which gives $$f(x)=-\frac{1}{2x^3}.$$ 6. Hence the correct option is $$\boxed{\text{C: } -\frac{1}{2x^3}}.$$
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