Integral from 0 to 1 of sqrt(x) * (x^2 + 1)dx

I know I have to substitute the sqrt x and the only way I've heard to do it is by substituting in trig, which I've never done before. Any help? Not entirely asking for an answer, just how I'm supposed to do this. Thanks :)

int (0, 1): sqrt(x) * (x^2 + 1) dx

No need to substitute here actually, just write sqrt(x) as x^(1/2):

int (0, 1): x^(1/2) * (x^2 + 1) dx

Now, distribute, remembering when you multiply powers they get added:

int (0, 1): x^(3/2) + x^(1/2) dx

This is a very simple integral to do via the power rule, no substitutions needed:

(2/5)x^(5/2) + (2/3)x^(3/2) | (0, 1)

Now just plug these in:

((2/5)(1)^(5/2) + (2/3)(1)^(3/2)) - ((2/5)(0)^(5/2) + (2/3)(0)^(3/2))

The entire second part is (0 + 0), and 1^(5/2) and 1^(3/2) are still 1, so the answer is just:

2/5 + 2/3 = 16/15

There's a much easier way than substitution. Just multiply out and use the power rule. Remember to consider techniques you have learned previously, and not just techniques you are learning now.

integral from 0 to 1 of sqrt(x) * (x^2 + 1) dx
= integral from 0 to 1 of (sqrt(x) * x^2 + sqrt(x)) dx
= integral from 0 to 1 of (x^(5/2) + x^(1/2)) dx
= [(2/7)x^(7/2) + (2/3)x^(3/2)] evaluated for x from 0 to 1
= 2/7 + 2/3 - (0 + 0)
= 20/21.

Lord bless you today!

∫[0,1] (x² + 1)√x dx

Let √x=sin(v), x=sin²(v), dx=d(sin²(v))=2sin(v)cos(v) dv and [0,1] → [0,π/2]
∫[0,π/2] (sin⁴(v) + 1)sin(v) *2sin(v)cos(v) dv =
=∫[0,π/2] 2(sin⁴(v) + 1)sin²(v)cos(v) dv =
=∫[0,π/2] 2(sin⁴(v) + 1)sin²(v) d(sin(v)) =
Let sin(v)=u and [0,π/2] → [0,1]
=∫[0,1] 2(u⁴ + 1)u² du=
=∫[0,1] 2u⁶ + 2u² du=
= 2u⁷/7 + ⅔u³ | [0,1] =
= 2/7 + ⅔=20/21

Int[sqrtx (x^2+1)]=
Int(x^2.5+x^0.5]=

1/3.5 (x^3.5) + 1/1.5 (x^1.5)= x^3.5/3.5+ x^1.5/1.5=

from zero to 1---------->1/3.5+1/1.5

ʃ(0 -> 1) sqrt(x) * (x^2 + 1)dx = ʃ(0 -> 1)(x^5/2 +x^1/2)dx =

= (0 -> 1)[(2/7)x^7/2 +1/2x^1/2] = [2/7 +1/2] - [0 +∞] =

= ∞ >=========================< ANSWER