Question

$\\ \dagger\ \; \bf Integral\ challenge\ :$
$\\ \displaystyle \bullet\ \; \red{\rm \int \dfrac{1}{x^2(x^4+1)^{\frac{3}{4}}}\ dx}$
$\\ \boxed{\underline{\bf Best\ of\ luck\ Guys\ ! }}$

Answer 1

Question :

$\displaystyle \dagger\ \; \sf \red{ \int \dfrac{1}{x^2(x^4+1)^{\frac{3}{4}}}\ dx}$

Solution :

$\displaystyle \sf \int \dfrac{1}{x^2(x^4+1)^{\frac{3}{4}}}\ dx$

$\longrightarrow \displaystyle \sf \int \dfrac{1}{x^2 \left[ x^4 \left( 1+ \frac{1}{x^4}\right) \right]^{\frac{3}{4}}}\ dx$

$\longrightarrow \displaystyle \sf \int \dfrac{1}{x^2.x^3 \left( 1+ \frac{1}{x^4}\right) ^{\frac{3}{4}}}\ dx$

$\longrightarrow \displaystyle \sf \int \dfrac{1}{x^5 \left( 1+ \frac{1}{x^4}\right) ^{\frac{3}{4}}}\ dx$

Let's use substitution method ,

$\implies \sf u=1 + \dfrac{1}{x^4}$

$\implies \sf du=- \dfrac{4}{x^5}\ dx$

$\implies \sf - \dfrac{du}{4} = \dfrac{1}{x^5}\ dx$

$\longrightarrow \displaystyle \sf \int \dfrac{-1}{4(u)^{\frac{3}{4}}}\ du$

$\longrightarrow \displaystyle \sf \dfrac{-1}{4} \int u^{-\frac{3}{4}}\ du$

$\longrightarrow \displaystyle \sf \dfrac{-1}{4} \left[ \dfrac{u^{-\frac{3}{4}+1}}{-\frac{3}{4}+1} \right]\ +c$

$\longrightarrow \displaystyle \sf \dfrac{-1}{4} \left[ \dfrac{u^{\frac{1}{4}}}{\frac{1}{4}} \right]\ +c$

$\longrightarrow \displaystyle \sf -u^{\frac{1}{4}} +c$

$\bullet\ \; \sf We\ have\ ,\ u=1+ \dfrac{1}{x^4}$

$\longrightarrow \displaystyle \sf \pink{-\left( 1+ \dfrac{1}{x^4}\right)^{\frac{1}{4}} +c}$

★ ═════════════════════ ★

Answer 2

$\displaystyle \bullet\ \; \purple{\rm \int \dfrac{1}{x^2(x^4+1)^{\frac{3}{4}}}\ dx}$

$\displaystyle \implies\; {\rm \int \dfrac{1}{x^2(x^4+1)^{\frac{3}{4}}}\ dx}$

$\displaystyle \implies\; {\rm \int \dfrac{1}{x^5\:\bigg(1+\dfrac{1}{x^4}\bigg)^{\frac{3}{4}}}\ dx}$

$\displaystyle \implies\; {\rm \int \dfrac{dx}{x^5}\:\dfrac{1}{\bigg(1+\dfrac{1}{x^4}\bigg)^{\frac{3}{4}}}}$

Let,

• $\sf{1\:+\:\dfrac{1}{x^4}\:=\:u}$

$\longrightarrow\:\sf{du\:=\:(-4x^{-5})\:.\:dx}$

$\longrightarrow\:\sf{du\:=\:\dfrac{-4}{x^{5}}\:.\:dx}$

$\longrightarrow\:\sf{\dfrac{du}{-4}\:=\:\dfrac{dx}{x^{5}}\:}$

☛ Now putting these value in the above equation given as,

$\displaystyle \implies\; {\rm \int \dfrac{du}{-4}\:\dfrac{1}{u^{\frac{3}{4}}}}$

$\displaystyle \implies\; {\rm \dfrac{1}{-4}\int \dfrac{du}{u^{\frac{3}{4}}}}$

$\displaystyle \implies\; {\rm \dfrac{1}{-4}\int u^{-\frac{3}{4}}\:du}$

$\displaystyle \implies\; {\rm \dfrac{1}{-4} \bigg[\dfrac{u^{-\frac{3}{4}\:+\:1}}{-\frac{3}{4}\:+\:1} \:+\:C\bigg]}$

$\displaystyle \implies\; {\rm \dfrac{1}{-4} \bigg[\dfrac{u^{\frac{1}{4}}}{\frac{1}{4}} \bigg]\:+\:C}$

$\displaystyle \implies\; {\rm \dfrac{1}{-4} \times{4} \: \bigg[u^{\frac{1}{4}}\bigg]\:+\:C}$

$\displaystyle \implies\; {\rm -1 \times{u^{\frac{1}{4}}}\:+\:C}$

$\displaystyle \implies\; {\rm -u^{\frac{1}{4}}\:+\:C}$

$\displaystyle \implies\; \green {\rm -\bigg(1\:+\:\dfrac{1}{x^4}\bigg)^{\frac{1}{4}}\:+\:C}$