Find the indefinite integral

Question

$$\int \frac{x^3 - 2x +1}{\sqrt{x}} dx$$

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?

Answer 1

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 $x^3=\frac{1}{4}x^4$, ...

Answer 2

Hint: $$ \int \frac{x^3 - 2x +1}{\sqrt{x}} dx = \int \frac{x^3}{\sqrt{x}} - \frac{2x}{\sqrt{x}} +\frac{1}{\sqrt{x}} dx = $$ $$ =\int x^{5/2}dx-2\int x^{1/2}dx+\int x^{-1/2}dx $$

Answer 3

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 $\frac{x^{\alpha+1}}{\alpha+1}$