Given `int_0^1root(3)(1+7x)dx`

Integrate using the u-substitution method.

Let `u=1+7x`

`(du)/dx=7`

`dx=(du)/7`

`=int_0^1u^(1/3)*(du)/7`

`=1/7int_0^1u^(1/3)du`

`=1/7*u^(4/3)/(4/3)` Evaluated from x=0 to x=1.

`=1/7*3/4*(1+7x)^(4/3)` Evaluated from x=0 to x=1.

`=3/28 [(1+7*1)^(4/3)-(1+7*0)^(4/3)]`

`=3/28[8^(4/3)-1^(4/3)]`

`=3/28[16-1]`

`=3/28[15]`

`=45/28`

`=1.607`

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