Evaluate: `int(1/(x3+x4)+(ln(1+x6)/(x3+sqrt(x)))dx`

Answer

`(3)/(2)x^(2//3)-(12)/(7)x^(7//12)+(4)/(3)x^(1//2)-(12)/(5)x^(5//12)+(1)/(2)x^(1//3)-4x^(1//4)-7x^(1//6)-12x^(1//12)+(2x^(1//2)-3x^(1//3)+6x^(1//6)+11)"In"(1+x^(1//6))+12"In"(1+x^(1//2))-3["In(1+x^(1//6))]^(2)+c`

Explanation

Let `I=int ((1)/(root(3)(x)+root(4)(x))+(In(1+root(6)(x)))/(root(3)(x)+sqrt(x)))dx`
`therefore" "I=I_(1)+I_(2)`
where, `I_(1)=int((1)/(root(3)(x)+root(4)(x)))dx`,
`I_(2)=int(In(1+root(6)(x)))/(root(3)(x)+root(4)(x))dx`
Now, `I_(1)=int((1)/(root(3)(x)+root(4)(x)))dx`
Put `x=t^(12) rArr dx = 12 t^(11)dt`
`therefore" "I_(1)=12 int(t^(11))/(t^(4)+t^(3t))d`
`" "=12 int(t^(8)dt)/(t+1)`
`" "=12 int(t^(7)-t^(6)+t^(5)-t^(4)+t^(3)-t^(2)+t-1)dt + 12 int(dt)/(t+1)`
`" "=12((t^(8))/(8)-(t^(7))/(7)+(t^(6))/(6)-(t^(5))/(5)+(t^(4))/(4)-(t^(3))/(3)+(t^(2))/(2)-t)+12 In (t+1)`
and `I_(2)=int{(In(1+root(6)(x)))/(root(3)x+sqrt(x))}dx`
Put `x = u^(6) rArr dx = 6 u^(5) du`
`therefore I_(2)=int(In(1+u))/(u^(2)+u^(3))6u^(5)du=int(In(1+u))/(u^(2)(1+u))*6u^(5)du`
`=6int(u^(3))/((u+1))In(1+u)du`
`=6int((u^(3)-1+1)/(u+1))In(1+u)du`
`=6int(u^(2)-u+1-(1)/(u+1))In(1+u)du`
`=6 int(u^(2)-u+1)In(1+u)du-6int(In(1+u))/((u+1))du`
`=6((u^(3))/(3)-(u^(2))/(2)+u)In(1+u)-int(2u^(3)-3u^(2)+6u)/(u+1)du-6(1)/(2)[In(1+u)]^(2)`
`=(2u^(3)-3u^(2)+6u)In(1+u)-int(2u^(2)-5u+(11u)/(u+1))du-3[In(1+u)]^(2)`
`=(2u^(3)-3u^(2)+6u)In(1+u)-((2u^(3))/(3)-(5)/(2)u^(2)+11u-11 In(u+1))-3[In(1+u)]^(2)`
`therefore" "I=(3)/(2)x^(2//3)-(12)/(7)x^(7//12)+2x^(1//2)-(12)/(5)x^(5//12)+3x^(1//3)-4x^(1//4)`
`-6x^(1//6)-12x^(1//12)+12 In(x^(1//12)+1)+(2x^(1//2)-3x^(1//3)+6x^(1//6))In(1+x^(1//6))`
`-[(2)/(3)x^(1//2)-(5)/(2)x^(1//3)11x^(1//6)-11 In(1+x^(1//6))]-3[In(1+x^(1//6))]+c`
`=(3)/(2)x^(2//3)-(12)/(7)x^(7//12)+(4)/(3)x^(1//2)-(12)/(5)x^(5//12)+(1)/(2)x^(1//3)-4x^(1//4)-7x^(1//6)-12x^(1//12)`
`+(2x^(1//2)-3x^(1//3)+6x^(1//6)+11)In(1+x^(1//6))+12 In (1+x^(1//12))-3[In(1+x^(1//6))]^(2)+c`