integration (1 Viewer)

[dawso]

gday all, got this one from cambridge, no idea what the hell they are doing.... anyone got a solution for it, cheers

int: x^3 / sqrt (x^2 + 1)

 

[dawso]

yeah thats rite, just delete ur reply once again....

 

[martin]

int(x^3 dx/sqrt(x^2+1))

let u^2=x^2+1 then 2u du/dx = 2x so u du= x dx

so int(x^3 dx/sqrt(x^2+1)
= int((u^2-1) u du / sqrt(u^2))
= int(u^2-1)du

you can take it from there, u=x^2+1 might work as well.

 

[KFunk]

You could also just do it in a single straight substitution:

∫ x³/√(x² + 1) dx

let u = x² +1 ----> du = 2xdx

int = 1/2∫ x²2x/√(x² + 1) dx

= 1/2∫ (u-1)/√u du

= 1/2[2/3.u^3/2 - 2√u] + c

= 1/3(x² +1)√(x² +1) - √(x² +1) + c

= 1/3.(x² - 2)√(x² +1) + c



EDIT: i.e the second one martin said