$$\int \frac{x^3 - 2x +1}{\sqrt{x}} dx$$
First term: $x^3 = \frac{1}{4}x^4$
Second term: $2x = x^2$
Third term: $1 = x$
Fourth term: $\sqrt{x} = x^{1/2}$
I know the fourth term is wrong and forget how to solve that one but which other ones are wrong as well?
You cannot do it this way. You need to divide each term in the numerator by the denominator first. Make it
$$x^{5/2}-2x^{1/2}+x^{-1/2}$$
Then integrate each term using power rule, which says
$$\int x^n dx=\frac{1}{n+1}x^{n+1} +C$$
Also mathematically, you cannot write $x^3=\frac{1}{4}x^4$, ...
Hint: $$ \int \frac{x^3 - 2x +1}{\sqrt{x}} dx = \int \frac{x^3}{\sqrt{x}} - \frac{2x}{\sqrt{x}} +\frac{1}{\sqrt{x}} dx = $$ $$ =\int x^{5/2}dx-2\int x^{1/2}dx+\int x^{-1/2}dx $$
write the integrand as $x^{\frac{5}{2}}-2x^{\frac{1}{2}}+x^{\frac{1}{2}}$ and use the fact that the antiderivative of $x^{\alpha}$ is $\frac{x^{\alpha+1}}{\alpha+1}$
