If dxx^3(1+x^6)^2/3=xf(x)(1+x^6)^1/3+C, where C is a constant of integration, then the fun…

Question

If $\displaystyle\int\frac{dx}{x^3(1+x^6)^{2/3}}=xf(x)(1+x^6)^{1/3}+C$, where C is a constant of integration, then the function $f(x)$ is equal to :

Accepted Answer

I = $$\int {{{dx} \over {{x^3}{{\left( {1 + {x^6}} \right)}^{{2 \over 3}}}}}} $$

= $$\int {{{dx} \over {{x^7}{{\left( {{1 \over {{x^6}}} + 1} \right)}^{{2 \over 3}}}}}} $$

Let $${{1 \over {{x^6}}} + 1}$$ = t

$$ \Rightarrow $$ $${{ - 6} \over {{x^7}}}dx = dt$$

$$ \Rightarrow $$ $${{dx} \over {{x^7}}} =  - {{dt} \over 6}$$

$$ \therefore $$ I = $$ - {1 \over 6}\int {{{dt} \over {{t^{{2 \over 3}}}}}} $$

= $$ - {1 \over 6}\left[ {{{{t^{{1 \over 3}}}} \over {{1 \over 3}}}} \right] + C$$

= $$ - {1 \over 2}{\left[ {{1 \over {{x^6}}} + 1} \right]^{{1 \over 3}}} + C$$

= $$ - {1 \over 2}{{{{\left( {1 + {x^6}} \right)}^{{1 \over 3}}}} \over {{x^2}}} \times {x \over x} + C$$

= $$x.\left( { - {1 \over {2{x^3}}}} \right).{\left( {1 + {x^6}} \right)^{{1 \over 3}}} + C$$

$$ \therefore $$ f(x) = $${ - {1 \over {2{x^3}}}}$$

Solutions · 1