Posted by **Tommy** on Thursday, April 19, 2012 at 9:55pm.

This is also one I can't figure out:

∫ x^3 (2x^2 + 1)^1/2 dx

u is obviously = 2x^2 + 1

du = 4x dx

x dx = 1/4 du

1/4 ∫ x^3 u^1/2

from here I know I can't integrate because I can't have more than 1 variable.

Thank you

- Integral calculus 2 -
**Reiny**, Thursday, April 19, 2012 at 11:31pm
I rewrote it as

∫ ( x(2x^2+1)^(1/2) (x^2) dx

I let u = x^2 and dv = x(2x^2 + 1)^(1/2) dx

du/dx = 2x

du = 2x dx

v = (1/6)(2x^2 + 1)^(3/2)

so we need: uv - ∫ v du

uv

= (1/6)(2x^2+1)^(3/2 (x^2)

= (x^2/6)(2x^2 + 1)^(3/2)

∫ (1/6)(2x^2 + 1)^(3/2) (2x) dx

= (1/30)(2x^2 + 1)^(5/2)

finally our integral

= (x^2/6)(2x^2 + 1)^(3/2) - (1/30)(2x^2 + 1)^(5/2)

= (1/30)(2x^2 + 1)^(3/2) [ 5x^2 - (2x^2 + 1) ]

= (1/30)(2x^2 + 1)^(3/2) (3x^2 - 1)

wow, you better check my steps on this one.

- Integral calculus 2 -
**Tommy**, Friday, April 20, 2012 at 10:46pm
thank you

## Answer This Question

## Related Questions

- Calculus 2 correction - I just wanted to see if my answer if correct the ...
- Calculus - Alright, I want to see if I understand the language of these two ...
- Calculus III - Use symmetry to evaluate the double integral ∫∫R(10+x...
- Integral Help - 1.) ∫ (sin x) / (cos^2 x) dx 2.) ∫ (1) / (1+x^2) dx ...
- Calculus - Evaluate the triple integral ∫∫∫_E (x+y)dV where E ...
- Calculus - Evaluate the triple integral ∫∫∫_E (x)dV where E is...
- Calculus - Find the integral by substitution ∫ [(16 x3)/(x4 + 5)] dx &#...
- Calculus - Evaluate the triple integral ∫∫∫_E (xy)dV where E ...
- Calculus - Evaluate the triple integral ∫∫∫_E (z)dV where E is...
- Integral Calculus - Trying to find ∫x*arctan(x)dx, but I can't figure out ...

More Related Questions