How to solve $\int x^{-1/4} \sqrt{1+x^{1/4}}$ using the substitution $u = 1 + x^2$

Question

I have this question in calculus but can't find a way to solve it the "intended way".

The question is this:

The integral ${\displaystyle \int x^{-1/4}\sqrt{1+x^{1/4}}} \ dx $ can be solved utilizing the substitution $u=1+x^2$. After the change of variables, we have an integral ${\displaystyle \int f(u) \ du}$ wich results in $F(u) +C$. Determine the primitive function $F(u)$.

Answer: $\dfrac{8}{7}u^{\frac{7}{2}}-\dfrac{16}{5}u^{\frac{5}{2}}+\dfrac{8}{3}u^{\frac{3}{2}}$

The thing is that I can't find a way to solve this using the given substitution.

I did find the answer by solving the integral using a diferent substituition $t=x^{1/4}$, but I realy don't know how to approach this exercise by utilizing the given substituition.

Answer 1

See what I have done.

Put $t^2 = x^1/4 +1$ and represent the original integral above in terms of $t$ and so you will get the integrand as $$8t^2(t^2-1)^2dt$$ and when you will integrate it and solve for the integral in terms of $t$ you will get the answer as $$8/7t^7-16/5t^3+8/3t^3+C$$ and as per you answer in terms of $u$ we must have $t^2=u$ but as per my substitution $t^2=x^1/4+1 =u$ but as per yours it is $u=1+x^2 $.
Therefore, your answer in terms of substitution in $u$ is wrong and final answer in terms of $u$ will be it you replace $t$ with $u=(t^2-1)^8+1$.

Thanks.

Answer 2

let $\sqrt{1+\sqrt[4]{x}}=t$.

Thus, $x=(t^2-1)^4$, $\frac{1}{4\sqrt[4]{x^3}}dx=2tdt$ and $$\int x^{-1/4}\sqrt{1+x^{1/4}}dx=\int x^{-1/4}t8\sqrt[4]{x^3}tdt=8\int t^2(t^2-1)^2dt=...$$