Vtyjkyuk

Find the indefinate integral with respect to $x$ of $$fracx^2-4x+10x^2sqrt x$$

For this problem I first made each individual number in the numerator separate from each other for easier integration and then simplified

$$=int left(fracx^2x^2sqrt x-frac4xx^2sqrt x+frac10x^2sqrt xright)dx$$
$$=int left(x^2x^-frac32-4x^1x^-frac32+10x^-frac32right)dx$$
$$=int left(x^frac12-4x^-frac12+10x^-frac32right)dx$$

I then integrated the expression to get

$$frac2x^frac323-8x^frac12-20x^-frac12$$

This is, however, wrong. Any ideas as to why?

Thanks in advance $:)$

$$fracx^2-4x+10x^2sqrt x$$
$$frac1sqrtx-frac4xsqrtx+frac10x^2sqrtx$$
$$x^-1/2-4x^-3/2+10x^-5/2$$

So it should be $x^-5/2$ instead of $x^-3/2$

$$I=int x^-1/2-4x^-3/2+10x^-5/2 dx=2x^1/2+8x^-1/2-frac203x^-3/2+C$$

Check again it should be

$$=int left(x^-frac12-4x^-frac32+10x^-frac52right)dx$$

indeed $frac1x^2sqrt x=x^-frac52$.

It is $x^-5/2$ and not $x^-3/2$.

HINT

Place $t=sqrtx$, after you solve the integral of a rational function.

You could have noticed that the $x^2$ simplify in the first term,

$$fracx^2+cdotsx^2sqrt x=frac1sqrt x+cdots=x^-1/2+cdots$$

So with a little more care,

$$int(x^-1/2-4x^-3/2+10x^-5/2),dx=frac21x^1/2-frac21(-4)x^-1/2-frac2310x^-3/2.$$