# `int x/sqrt(6x+1) dx` Find the indefinite integral

`intx/sqrt(6x+1)dx`

Apply integral substitution: `u=6x+1`

`=>(u-1)=6x`

`=>x=(u-1)/6`

`dx=1/6(du)`

`intx/sqrt(6x+1)dx=int((u-1)/6)/sqrt(u)(1/6)du`

`=int1/36(u-1)/(sqrt(u))du`

Take the constant out,

`=1/36int(u-1)/sqrt(u)du`

`=1/36int(u/sqrt(u)-1/sqrt(u))du`

`=1/36int(u^(1/2)-u^(-1/2))du`

Apply the sum rule,

`=1/36{intu^(1/2)du-intu^(-1/2)du}`

Apply the power rule,

`=1/36{(u^(1/2+1)/(1/2+1))-(u^(-1/2+1)/(-1/2+1))}`

`=1/36{u^(3/2)/(3/2)-u^(1/2)/(1/2)}`

`=1/36{2/3u^(3/2)-2u^(1/2)}`

Substitute back `u=(6x+1)` and add a constant C to the solution,

`=1/36(2/3(6x+1)^(3/2)-2(6x+1)^(1/2))+C`

`=1/36(2)(6x+1)^(1/2)(1/3(6x+1)-1)+C`

`=1/18sqrt(6x+1)((6x+1-3)/3)+C`

`=sqrt(6x+1)/18((6x-2)/3)+C`

`=sqrt(6x+1)/18(2/3)(3x-1)+C`

`=1/27(3x-1)sqrt(6x+1)+C`

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`intx/sqrt(6x+1)dx`

Apply integral substitution: `u=6x+1`

`=>(u-1)=6x`

`=>x=(u-1)/6`

`dx=1/6(du)`

`intx/sqrt(6x+1)dx=int((u-1)/6)/sqrt(u)(1/6)du`

`=int1/36(u-1)/(sqrt(u))du`

Take the constant out,

`=1/36int(u-1)/sqrt(u)du`

`=1/36int(u/sqrt(u)-1/sqrt(u))du`

`=1/36int(u^(1/2)-u^(-1/2))du`

Apply the sum rule,

`=1/36{intu^(1/2)du-intu^(-1/2)du}`

Apply the power rule,

`=1/36{(u^(1/2+1)/(1/2+1))-(u^(-1/2+1)/(-1/2+1))}`

`=1/36{u^(3/2)/(3/2)-u^(1/2)/(1/2)}`

`=1/36{2/3u^(3/2)-2u^(1/2)}`

Substitute back `u=(6x+1)` and add a constant C to the solution,

`=1/36(2/3(6x+1)^(3/2)-2(6x+1)^(1/2))+C`

`=1/36(2)(6x+1)^(1/2)(1/3(6x+1)-1)+C`

`=1/18sqrt(6x+1)((6x+1-3)/3)+C`

`=sqrt(6x+1)/18((6x-2)/3)+C`

`=sqrt(6x+1)/18(2/3)(3x-1)+C`

`=1/27(3x-1)sqrt(6x+1)+C`

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