# How do you use substitution to integrate #x(sqrt(2x+1))dx.#?

##### 2 Answers

This is better-solved using integration by parts. There is nothing u-substitution-related that is easy to think of quickly, here.

I would choose the function that is easier to differentiate and make go away as

Let:

#u = x# #du = dx# #dv = (2x+1)^(1//2)dx# #v = (2//3)/2(2x+1)^(3//2) = 1/3(2x+1)^(3//2)#

Then, we get:

#int udv = uv - intvdu#

#= x/3(2x+1)^(3//2) - int1/3(2x+1)^(3//2)dx#

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

#= x/3(2x+1)^(3//2) - 1/15(2x+1)^(5//2)#

#= (2x+1)^(3//2)[x/3 - 1/15(2x+1)]#

#= 1/15(2x+1)^(3//2)[5x - (2x+1)]#

And now we tack on the constant at the end.

#= color(blue)(ul(1/15(2x+1)^(3//2)(3x-1) + C))#

Please see below.

#### Explanation:

Let

So

# = 1/2 int 1/2(u-1)u^(1/2)du#

# = 1/4 int (u^(3/2)-u^(1/2)) du#

# = 1/4 (2/5u^(5/2) - 2/3u^(3/2)) +C#

# = 1/10u^(5/2)-1/6u^(3/2)+C#

Simplify as desired before reversing the substitution.

I like

So we have

# = sqrt(2x+1)/30 (2x+1) (6x-2) +C#

# = sqrt(2x+1)/15 (2x+1) (3x-1) +C#