# Using a suitable substitution evaluate ` `` ``int_1^8(1/(x^(4/3)+x^(2/3)))dx` `` ` ````.` ` <br> ` ` `

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jeew-m | Certified Educator

Let us take `x = t^3` as the substitution.

`x = t^3`

`dx = 3t^2dt`

When x = 1 then t = 1

When x = 8 then t = 2

`int_1^8 1/(x^(4/3)+x^(2/3))dx`

`= int_1^2(1/(t^4+t^2))3t^2dt`

`= 3int_1^2(1/(1+t^2))dt`

`= 3[tan^(-1)t]_1^2`

`= 3(tan^(-1)2-tan^(-1)1)`

`= 3tan^(-1)2-(3pi)/4`

** So the answer is **`3tan^(-1)2-(3pi)/4`

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