Integration involving roots $$\int\frac{dx}{(1+x^\frac{1}{4})x^\frac{1}{2}}$$
This is my work:
$$u^4=x$$
$$4u^3=dx$$
$$\int\frac{4u^3du}{(1+u)u^2}=\int\frac{4u^3du}{(1+u)u^2}=-4(1+x^\frac{1}{4})^{-1}+2(1+x^\frac{1}{4})^{-2}+C$$
But wolframAlpha gives quite a different answer. So, have I done a mistake? Or somehow both answers are correct?
 A: You substitutions are just fine. First, we cancel the common factor of $u^2$ in the numerator and denominator.
$$\int\frac{4u^3du}{(1+u)u^2}=\int\frac{4u\,du}{1+u}$$
Dividing the numerator by the denominator gives us $$\int\frac{4u}{1+u}\,du = \int \left(4 - \frac 4{1+u} \right) \,du = 4\int \,du - 4\int \frac {du}{1+u}\tag{1}$$
Now, I'm not entirely sure how you obtained your integrated your terms, but I can only guess that you mixed up differentiating and integrating terms: subtracting 1 from the exponent of $(1+u)^0$ and dividing $4$ by $-1$, and subtracting 1 from the exponent of $(1+u)^{-1}$ and dividing $-4$ by $-2$. 
First, when integrating a constant $\alpha$, $\int \alpha \,du$, we obtain the variable with respect to which we are integrating, and in this case, that's simply the variable $u$ (the $\,du\,$ indicates which variable). So we can think of $$\int \alpha\,du = \int \alpha u^0\,du = \alpha u^1 + C = \alpha u + C$$ Second, when integrating, we add (increment) exponents, and divide by the new exponent, except when the exponent is $-1$ (and in your second integral, we have $(1 + u)^{-1}$ as the integrand. Recall, $$\int \frac{f'(u)}{f(u)}\,du = \ln|f(u)| + C.$$
So back to $(1)$. Integrating, we get $$4\int \,du - 4\int \frac {du}{1+u} = 4u - 4\ln|(1+u)| + C = 4x^{1/4} - 4 \ln\left(1 + x^{1/4}\right) + C$$
A: Check this calculations. They give the same result as wolfram alpha
$$
\int\frac{4u}{1+u}du=
4\int\left(1-\frac{1}{u+1}\right)du=
4\int du-4\int\frac{1}{u+1}du=
4u-4\ln(1+u)=\\
4x^{1/4}-4\ln(1+x^{1/4})
$$
