# Integral of 1/ sqrt ( 1+ cubicroot(x)) dx

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You need to evaluate the indefinite integral `int 1/(sqrt(1 + root(3) x)) dx` , using the following substitution `1 + root(3) x = t` , such that:

`int 1/(sqrt(1 + root(3) x)) dx`

`1 + root(3) x = t => root(3) x = t - 1 => (1/(3root(3)(x^2))) dx = dt`

`dx = 3(t - 1)^2dt`

`int 1/(sqrt(1 + root(3) x)) dx = int (3(t - 1)^2)/(sqrt t) dt`

You need to expand the binomial `(t - 1)^2 = t^2 - 2t + 1` , such that:

`int (3(t - 1)^2)/(sqrt t) dt = 3int (t^2 -2t + 1)/(sqrt t) dt`

Using the property of linearity of integral, yields:

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

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

`3int (t^2 -2t + 1)/(sqrt t) dt = 3 t^(3/2 + 1)/(3/2 + 1) - 6*t^(3/2)/(3/2) + 6 sqrt t + c`

`3int (t^2 -2t + 1)/(sqrt t) dt = (6/5)*t^2sqrt t - 4t*sqrt t + 6sqrt t + c`

Replacing back `1 + root(3) x` for t yields:

`int 1/(sqrt(1 + root(3) x)) dx = (6/5)*(1 + root(3) x)^2sqrt(1 + root(3) x) - 4(1 + root(3) x)*sqrt (1 + root(3) x) + 6sqrt (1 + root(3) x) + c`

Hence, evaluating the indefinite integral yields `int 1/(sqrt(1 + root(3) x)) dx = (6/5)*(1 + root(3) x)^2sqrt(1 + root(3) x) - 4(1 + root(3) x)*sqrt (1 + root(3) x) + 6sqrt (1 + root(3) x) + c.`

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