Question

Question #ae4e5

Answer 1

`I=2x^(1/2)-3x^(1/3)+6x^(1/6)-6ln|x^(1/6)+1|+C.`

Explanation:

Let `I=int1/(x^(1/2)+x^(1/3))dx`

Observe that, the lcm of `2 and 3` is `6`, so, we select the

substitution `x=t^6," giving, "dx=6t^5dt, x^(1/2)=t^3, x^(1/3)=t^2`.

`:. I=int(6t^5)/(t^3+t^2)dt=6intt^3/(t+1)dt=6int(t^3+1-1)/(t+1)dt`

`=6int[(t^3+1)/(t+1)-1/(t+1)]dt=6int[t^2-t+1-1/(t+1)]dt`

`=6{t^3/3-t^2/2+t-ln|t+1|}`

`=2t^3-3t^2+6t-6ln|t+1|,` and as `t=x^(1/6)`,

`:. I=2x^(1/2)-3x^(1/3)+6x^(1/6)-6ln|x^(1/6)+1|+C.`

Enjoy Maths.!