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

Question

Emilia Aguilar · Accepted Answer

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 = sqrt5tantheta#
#sqrt(x^4 + 5) = sqrt5sec^2theta#
#x = 5^(1/4)sqrttantheta#
#dx = 5^(1/4)*1/(2sqrttantheta)sec^2thetad theta = 5^(1/4)/2sec^2theta/(sqrttantheta)d theta# #= int sqrt5tantheta sqrt5sec^2theta 5^(1/4)/2sec^2theta/(sqrttantheta)d theta# #= 5^(5/4)/2int (tantheta)^(1/2) sec^4thetad theta# #= 5^(5/4)/2int (tantheta)^(1/2) (tan^2theta + 1)sec^2thetad theta# since #1+tan^2theta = sec^2theta#. Then just do some u-substitution with #u = tantheta# and #du = sec^2thetad 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).

Cameron Alexander · Answer

undefined To integrate $x^2\sqrt{x^4 + 5}$ using substitution, let $u = x^4 + 5$. Then, $du = 4x^3 dx$. Rearranging gives $x^3 dx = \frac{1}{4} du$. Now, substitute:

$$u = x^4 + 5$$
$$du = 4x^3 dx$$
$$x^3 dx = \frac{1}{4} du$$

So, the integral becomes:

$$\int x^2\sqrt{x^4 + 5} \, dx = \int \frac{1}{4} \sqrt{u} \, du$$

Now, you can integrate $\frac{1}{4} \sqrt{u}$ with respect to $u$. Once you find the antiderivative, don't forget to substitute back $x$ for $u$.