Integrate {x3+1/(whole root over)x2+x}dx

[perfectibilis]

Integrate the following--->
{x³+1/(whole root over)x²+x}dx

 

[foxjwill]

Do you mean [tex]\int \frac{x^3+1}{\sqrt{x^2+x}}\,dx[/tex] or [tex]\int \left(x^3+\frac{1}{\sqrt{x^2+x}}\right)dx?[/tex]

Either way, the best thing to do is to start by completing the square in the denominator, then using a bunch of trig substitution stuff.

 

[Дьявол]

x³+1³=(x+1)(x²-x+1)
x²+x=x(x+1)

Is it enough help?

 

[Mark44]

x³+1³=(x+1)(x²+x+1)
x²+x=x(x+1)

Actually, x³ + 1 = (x + 1)(x² - x + 1).

In any case, we still don't know exactly what the integrand is.

 

[perfectibilis]

Do you mean [tex]\int \frac{x^3+1}{\sqrt{x^2+x}}\,dx[/tex] or [tex]\int \left(x^3+\frac{1}{\sqrt{x^2+x}}\right)dx?[/tex]

Either way, the best thing to do is to start by completing the square in the denominator, then using a bunch of trig substitution stuff.

I mean the first image.

 

[Дьявол]

Mark44 thanks for the correction.

perfectibilis start by writing x³+1 with

[tex]\sqrt{(x+1)^2(x^2-x+1)^2}[/tex]