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

Jul 2, 2015

If I rewrite this as:

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:
${x}^{2} = \sqrt{5} \tan \theta$
$\sqrt{{x}^{4} + 5} = \sqrt{5} {\sec}^{2} \theta$
$x = {5}^{\frac{1}{4}} \sqrt{\tan} \theta$
$\mathrm{dx} = {5}^{\frac{1}{4}} \cdot \frac{1}{2 \sqrt{\tan} \theta} {\sec}^{2} \theta d \theta = {5}^{\frac{1}{4}} / 2 {\sec}^{2} \frac{\theta}{\sqrt{\tan} \theta} d \theta$

$= \int \sqrt{5} \tan \theta \sqrt{5} {\sec}^{2} \theta {5}^{\frac{1}{4}} / 2 {\sec}^{2} \frac{\theta}{\sqrt{\tan} \theta} d \theta$

$= {5}^{\frac{5}{4}} / 2 \int {\left(\tan \theta\right)}^{\frac{1}{2}} {\sec}^{4} \theta d \theta$

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

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

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

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