How do you find the integral of $\frac{d x}{(x^{\frac{5}{2}} \cdot {(1 + x^{2})}^{\frac{1}{4}})}$ $dx/((x^(5/2) * (1+ x^2)^(1/4))$ ?

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How do you find the integral of $\frac{d x}{(x^{\frac{5}{2}} \cdot {(1 + x^{2})}^{\frac{1}{4}})}$ $dx/((x^(5/2) * (1+ x^2)^(1/4))$ ?

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Answer 1 · 2018-07-21T17:35:42

$\int \frac{1}{x^{\frac{5}{2}} {(1 + x^{2})}^{\frac{1}{4}}} d x = - \frac{2}{3} {(\frac{1}{x^{2}} + 1)}^{\frac{3}{4}} + c$ $int1/(x^(5/2)(1+x^2)^(1/4))dx=-2/3(1/x^2+1)^(3/4)+c$

Explanation:

Here ,

$I = \int \frac{1}{x^{\frac{5}{2}} {(1 + x^{2})}^{\frac{1}{4}}} d x$ $I=int1/(x^(5/2)(1+x^2)^(1/4))dx$

Substitute $x = \frac{1}{t} \Rightarrow d x = - \frac{1}{t^{2}} d t$ $color(red)(x=1/t=>dx=-1/t^2dt$

So,

$I = \int \frac{1}{{(\frac{1}{t})}^{\frac{5}{2}} {(1 + \frac{1}{t^{2}})}^{\frac{1}{4}}} (- \frac{1}{t^{2}}) d t$ $I=int1/((1/t)^(5/2)(1+1/t^2)^(1/4))(-1/t^2)dt$

$= - \int \frac{t^{\frac{5}{2}}}{\frac{{(t^{2} + 1)}^{\frac{1}{4}}}{t^{\frac{1}{2}}}} \cdot \frac{1}{t^{2}} d t$ $=-intt^(5/2)/((t^2+1)^(1/4)/t^(1/2))*1/t^2dt$

$= - \int \frac{t^{\frac{5}{2} + \frac{1}{2} - 2}}{{(t^{2} + 1)}^{\frac{1}{4}}} d t$ $=-intt^(5/2+1/2-2)/(t^2+1)^(1/4)dt$

$:.I=-intt/(t^2+1)^(1/4)dt$

Substitute $color(blue)((t^2+1)^(1/4)=u=>t^2+1=u^4$

$=>2tdt=4u^3du=>tdt=2u^3du$

$I=-int(2u^3)/udu$

$=>I=-2intu^2du$

$=>I=-2[u^3/3]+c$

$=-2/3(u)^3+c$

Subst. back , $color(blue)(u=(t^2+1)^(1/4)$

$I=-2/3(t^2+1)^(3/4)+c$

Again subst. $color(red)(t=1/x$

$:.I=-2/3(1/x^2+1)^(3/4)+c$

$=>I=-2/3(1+x^2)^(3/4)/(x^2)^(3/4)+c$

$:.I=-(2(1+x^2)^(3/4))/(3x^(3/2))+c$

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How do you find the integral of $\frac{d x}{(x^{\frac{5}{2}} \cdot {(1 + x^{2})}^{\frac{1}{4}})}$ $dx/((x^(5/2) * (1+ x^2)^(1/4))$ ?

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Explanation:

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