How do i ∫x √2x+1 ?

Answer 1

let u = 2x+1
du = 2dx
du/2 = dx
and u = 2x+1
.... u- 1=2x
.... (u-1)/2 = x
int xsqrt(2x+1)dx= int ((u-1)/2 *u^(1/2) du/2
1/4 int u^3/2 -u^1/2 du
1/4 [ (2/5)u^(5/2) - (2/3)*u^(3/2)] +C
1/10 u^(5/2) - 1/6u^(3/2)+C
replace u = 2x+1
answer (1/10 )*(2x+1)^(5/2) - (1/6)* (2x+1)^(3/2)+C
good luck
watch out others 2 answers they got wrong . i am pretty sure. except they edited

Answer 2

I = ∫ x (2x + 1)^(½) dx
I = ∫ u (dv/dx) dx = uv - ∫ v (du/dx) dx
u = x and dv/dx = (2x + 1)^(1/2)
du/dx = 1
v = [ (1/2) (2x + 1)^(3/2) / (3/2) ]
v = (1/3) (2x + 1)^(3/2)
I = (x/3)(2x + 1)^(3/2) - ∫(1/3)(2x + 1)^(3/2)dx
I = (x/3)(2x + 1)^(3/2) - (1/3)(1/5) (2x + 1)^(5/2)
I = (2x + 1)^(3/2) [ x/3 - (1/15)(2x + 1) ]
I = (1/15)(2x + 1)^(3/2)[ 5x - 2x - 1]
I = (1/15)(2x + 1)^(3/2) (3x - 1) + C

Answer 3

int ∫x √(2x+1)dx=1/2*int x√(2x+1)d(2x+1)=
u=x; √(2x+1)d(2x+1)=dv
v=2/3*(2x+1)^3/2; du=dx
=1/2*[2/3*x(2x+1)^1,5-
-2/3*1/2*int (2x+1)^3/2*d(2x+1)]=
=x/3*(2x+1)^3/2-
-1/15*(2x+1)^5/2=
=1/3*(2x+1)^3/2*
*[x-1/5*(2x+1)]+c

Answer 4

∫x √(2x+1) dx ........I assume

u=2x+1
=> du = 2dx
=> 1/2 du = dx
&
2x+1 = u
=> x = ( u-1)/2

∫x √(2x+1) dx = ∫ (( u-1)/2)( u^1/2 ) 1/2du
=> 1/4 ∫ (u^3/2 - 1/2 u^1/2) du
=> 1/4 [ ( u^5/2)/(5/2) - (1/2 u^3/2)/(3/2)]
=> 1/4 [ 2/5 u^5/2 - 1/3 u^3/2 ]
=> 1/4 [ 2/5 (2x+1) ^5/2 - 1/3 (2x+1)^3/2 ] +c