Question

Could somebody please help me solve this integral?

`int1/(sqrtx*(root4x + root6x))dx`

Answer 1

`4root4x-6root6x+12root12x-12ln|root12x+1|+C.`

Explanation:

Observe that, the l.c.m. of `2,4 and 6` is `12`, so, let us substitute

`x=y^12,"so that, "dx=12y^11dy.`

Hence, `I=int1/{sqrtx(root4x+root6x)}dx`

`=int(12y^11)/{y^6(y^3+y^2)}dy=12inty^3/(y+1)dy`

`=12int{(y^3+1)-1}/(y+1)dy`

`=12int{(y^3+1)/(y+1)-1/(y+1)}dy`

`=12int{(y+1)(y^2-y+1)}/(y+1)dy-12int1/(y+1)dy`

`=12int(y^2-y+1)dy-12ln|y+1|`

`=12(y^3/3-y^2/2+y)-12ln|y+1|`

`=4y^3-6y^2+12y-12ln|y+1|, and, because, y=x^(1/12)`,

`=4x^(1/4)-6x^(1/6)+12x^(1/12)-12ln|x^(1/12)+1|`.

`:. I=4root4x-6root6x+12root12x-12ln|root12x+1|+C.`

Spread the Joy of Maths.!

Answer 2

`int (dx)/(sqrt(x)(root(4)x+root(6)x)) = 4(1+x^(1/12))^3-18(1+x^(1/12))^2+16+16x^(1/12)-12 ln abs (1+x^(1/12)) +C`

Explanation:

Simplify the expression of the integrand function:

`int (dx)/(sqrt(x)(root(4)x+root(6)x)) = int (dx)/(x^(1/2)(x^(1/4)+x^(1/6)) )= int (dx)/(x^(1/2)x^(1/6)(x^(1/4-1/6)+1) )= int(dx)/(x^(2/3)(1+x^(1/12))`

Substitute:

`t= x^(1/12)`

`x= t^12`

`dx = 12t^11dt`

`int (dx)/(sqrt(x)(root(4)x+root(6)x)) = int (12t^11dt)/(t^8(1+t)) = 12 int (t^3dt)/(1+t)`

Substitute:

`u= 1+t`

`du = dt`

`int (dx)/(sqrt(x)(root(4)x+root(6)x)) = int (12t^11dt)/(t^8(1+t)) = 12 int ((u-1)^3du)/u = 12 int (u^3-3u^2+3u-1)/u du = 12 int (u^2-3u+3-1/u) du = 4u^3-18u^2+36u-12lnabs(u)+C`

Substituting back:

`u = 1+t = 1+x^(1/12)`

`int (dx)/(sqrt(x)(root(4)x+root(6)x)) = 4(1+x^(1/12))^3-18(1+x^(1/12))^2+16+16x^(1/12)-12 ln abs (1+x^(1/12)) +C`

You can then develop and simplify the powers to have a more compact expression.