# $\int \frac{x^2-x}{x+\sqrt {x}}\, \mathrm{d}x$

My problem, is how to find the integral of $$\int \frac{x^2-x}{x+\sqrt {x}}\, \mathrm{d}x$$

And I don't even know where to start. I tried factoring out x, but I don't get anything useful. I need some suggestions please.

• Factor the top, using difference of perfect squares. Jan 5, 2014 at 9:44
• To integrate an integral, you first need to find the primitive of the integrand, and once found find a primitive of the resulting function. This is probably not what you meant. Moral: you should either say "find the integral" or "integrate the expression", but not what you said in the title and wrote in the question. Jan 5, 2014 at 9:48
• $x^2-x=(x-\sqrt x)(x+\sqrt x)$ Jan 5, 2014 at 9:49

Observe that $x^2-x=(x+\sqrt{x})(x-\sqrt{x})$

$$\int \frac{x^2-x}{x+\sqrt {x}}\, \mathrm{d}x= \int \frac{(x-1)\sqrt {x}}{1+\sqrt {x}}\, \mathrm{d}x=\int \left(\sqrt {x}-1\right)\sqrt {x}\, \mathrm{d}x$$

We have $$\int \frac{x^2-x}{x+\sqrt {x}}\, dx=\int (x-\sqrt{x} ) dx.$$

It is sum of table integrals.

Let $t=\sqrt x$ then $dx=2tdt$ hence $$\int \frac{x^2-x}{x+\sqrt {x}}\, \mathrm{d}x=2\int\frac{t^4-t^2}{t+1}\, \mathrm{d}t=2\int t^2(t-1)\, \mathrm{d}t$$ which we can integrate easily.

• Partial fraction decomposition for a denominator of degree$~1$, seriously? Jan 5, 2014 at 9:51
• Well in this case no, but in general why not? how you integrate $\int \frac{t^2}{t+1}dt$?
– user63181
Jan 5, 2014 at 9:55
• There was no any partial fraction decomposition, usual division
– Leox
Jan 5, 2014 at 9:56
• @Leox This usual division is a partial fraction decomposition.
– user63181
Jan 5, 2014 at 9:57
• @Sami Ben Romdhane, No. After this division we have got a polynomial, not a fraction.
– Leox
Jan 5, 2014 at 10:00

Factor the top using $a^2-b^2=(a+b)(a-b)$: $$\int \frac{(x+\sqrt{x})(x-\sqrt{x})}{x+\sqrt {x}}\, \mathrm{d}x$$ Simplifying we obtain: $$\int x-\sqrt{x}, \mathrm{d}x$$ Which can be easily integrated to: $$\frac{x^2}{2}-\frac{2x^{3/2}}{3}+c$$

If you didn't see that the fraction simplifies, you can always integrate a rational expression involving only a single simple square root by changing variable to that square root. Here putting $u=\sqrt x$ gives $x=u^2$ so $\def\d{\,\mathrm d}\!\d x=2u\d u$, and you get $$\int \frac{(x^2-x)\d x}{x+\sqrt {x}}=\int\frac{(u^4-u^2)2u\d u}{u^2+u}.$$ Now you are forced to do the polynomial division and find that $2u(u^4-u^2)=2(u^2+u)(u^3-u^2)$, so you are looking at $$\int2(u^3-u^2)\d u=\frac14u^4-\frac23u^3+C=\frac14x^2-\frac23x\sqrt x+C.$$