# What is `int sqrt(3x + 4) dx` ?

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We have to find `int sqrt(3x + 4) dx`

Let 3x + 4 = y

dy/dx = 3

=> dy/3 = dx

`int sqrt (3x+ 4) dx`

=> `(1/3) int sqrt y dy`

=> (1/3)*y^(3/2)/(3/2)

=> (2/9)*y^(3/2)

substitute y = 3x + 4

=> (2/9)*(3x + 4)^(3/2)

**The required integral is (2/9)*(3x + 4)^(3/2)**