Find the integral $\int \:x^{-\frac{1}{2}}\cdot \left(1+x^{\frac{1}{4}}\right)^{-10} dx$

Question

Help me find the integral. I think we have to somehow replace apply.

$$\int \:x^{-\frac{1}{2}}\cdot \left(1+x^{\frac{1}{4}}\right)^{-10} dx =\int \frac{1}{\sqrt{x} (1+x^{\frac{1}{4}})^{10}} dx $$

Answer 1

You may just perform the change of variable $x=u^4$, to get $$ \begin{align} \int \:x^{-\frac{1}{2}}\cdot \left(1+x^{\frac{1}{4}}\right)^{-10} dx&=4\int \:u^{-2}\cdot \left(1+u\right)^{-10} u^3du\\\\ &=4\int \frac{1}{(1+u)^9} du-4\int \frac{1}{(1+u)^{10}}du\\\\ &=-\frac{1}{2 (1+u)^8}+\frac{4}{9 (1+u)^9}+C\\\\ &=-\frac{1}{2 \left(1+x^{1/4}\right)^9}+\frac{4}{9 \left(1+x^{1/4}\right)^9}+C. \end{align} $$

Answer 2

Note that if $u = 1 + x^{1/4}$, then $du = (1/4)x^{-3/4}dx$. This becomes a helpful substitution if you replace $x^{-1/2}$ with $x^{-3/4}x^{1/4} = x^{-3/4}(u-1)$.

Answer 3

let $$x=u^4$$ $$dx=4u^3 du$$ $$\int \frac{4u^3 du}{u^2(1+u)^{10}}=\int (\frac{4}{(u+1)^9}-\frac{4}{(u+1)^{10}})du$$