How do you use substitution to integrate $x^{2} \sqrt{x^{4} + 5}$ $x^2sqrt(x^(4)+5)$ ?

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How do you use substitution to integrate $x^{2} \sqrt{x^{4} + 5}$ $x^2sqrt(x^(4)+5)$ ?

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Answer 1 · 2015-07-02T01:57:27

If I rewrite this as:

$x^{2} \sqrt{{(x^{2})}^{2} + {(\sqrt{5})}^{2}} \propto u \sqrt{u^{2} + a^{2}}$ $x^2 sqrt((x^2)^2 + (sqrt5)^2) prop u sqrt(u^2 + a^2$

then you can do the trig substitution method of letting:
$x^{2} = \sqrt{5} tan θ$ $x^2 = sqrt5tantheta$
$\sqrt{x^{4} + 5} = \sqrt{5} {sec}^{2} θ$ $sqrt(x^4 + 5) = sqrt5sec^2theta$
$x = 5^{\frac{1}{4}} \sqrt{tan θ}$ $x = 5^(1/4)sqrttantheta$
$d x = 5^{\frac{1}{4}} \cdot \frac{1}{2 \sqrt{tan θ}} {sec}^{2} θ d θ = \frac{5^{\frac{1}{4}}}{2} \frac{{sec}^{2} θ}{\sqrt{tan θ}} d θ$ $dx = 5^(1/4)*1/(2sqrttantheta)sec^2thetad theta = 5^(1/4)/2sec^2theta/(sqrttantheta)d theta$

$= \int \sqrt{5} tan θ \sqrt{5} {sec}^{2} θ \frac{5^{\frac{1}{4}}}{2} \frac{{sec}^{2} θ}{\sqrt{tan θ}} d θ$ $= int sqrt5tantheta sqrt5sec^2theta 5^(1/4)/2sec^2theta/(sqrttantheta)d theta$

$= \frac{5^{\frac{5}{4}}}{2} \int {(tan θ)}^{\frac{1}{2}} {sec}^{4} θ d θ$ $= 5^(5/4)/2int (tantheta)^(1/2) sec^4thetad theta$

$= \frac{5^{\frac{5}{4}}}{2} \int {(tan θ)}^{\frac{1}{2}} ({tan}^{2} θ + 1) {sec}^{2} θ d θ$ $= 5^(5/4)/2int (tantheta)^(1/2) (tan^2theta + 1)sec^2thetad theta$

since $1 + {tan}^{2} θ = {sec}^{2} θ$ $1+tan^2theta = sec^2theta$ .

Then just do some u-substitution with $u = tan θ$ $u = tantheta$ and $d u = {sec}^{2} θ d θ$ $du = sec^2thetad theta$ and you'll just have some polynomials to work with, integrate that, substitute back in previous variables, add $+ C$ $+C$ .

I think you can do it from there (I'm in a hurry. If someone wants to finish this, go ahead).

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How do you use substitution to integrate $x^{2} \sqrt{x^{4} + 5}$ $x^2sqrt(x^(4)+5)$ ?

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How do you use substitution to integrate $x^{2} \sqrt{x^{4} + 5}$ $x^2sqrt(x^(4)+5)$ ?