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Question

Mathematics Question on Integrals of Some Particular Functions

If dxx3(1+x6)23=f(x)(1+x6)13+C\int \frac{dx}{x^{3}\left(1+x^{6}\right)^{\frac{2}{3}}}=f \left(x\right)\left(1+x ^{6}\right)^{\frac{1}{3}}+C, where C is a constant of integration, then the function f(x)f \left(x\right) is equal to-

A

16x3- \frac{1}{6x^{3}}

B

3x2 \frac{3}{x^{2}}

C

12x2- \frac{1}{2x^{2}}

D

12x3- \frac{1}{2x^{3}}

Answer

12x3- \frac{1}{2x^{3}}

Explanation

Solution

dxx3(1+x6)23=xf(x)(1+x6)13+c\int \frac{dx}{x^{3}\left(1+x^{6}\right)^{\frac{2}{3}}}= x f \left(x\right)\left(1+x^{6}\right)^{\frac{1}{3}} +c
dxx7(1x6+1)23=xf(x)(1+x6)13+c\int \frac{dx}{x^{7}\left(\frac{1}{x^{6}}+1\right)^{\frac{2}{3}}}= x f \left(x\right)\left(1+x^{6}\right)^{\frac{1}{3}} +c
Let t=1x6+1t=\frac{1}{x^{6}}+1
dt=6x7dxdt=\frac{-6}{x^{7}}dx
=16dtt23=12t13=- \frac{1}{6}\int \frac{dt}{t^{\frac{2}{3}}}=- \frac{1}{2} t^{\frac{1}{3}}
=12(1x6+1)13=12(1+x6)13x2=- \frac{1}{2} \left(\frac{1}{x^{6}}+1\right)^{\frac{1}{3}} =- \frac{1}{2} \frac{\left(1+x^{6}\right)^{\frac{1}{3}}}{x^{2}}
f(x)=12x3\therefore f \left(x\right)=- \frac{1}{2x^{3}}