# What is the indefinite integral of the function y=(4x^3+1)square root(x^4+x)?

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### 2 Answers

We have to find the indefinite integral of y=(4x^3+1)*sqrt(x^4+x)

Int[(4x^3+1)*sqrt(x^4+x) dx]

let x^4 + x = t, dt/dx = 4x^3 + 1

=> Int[(sqrt t) dt]

=> t^(1/2 + 1)/(1/2 + 1)

=> t^(3/2)/(3/2)

=> (2/3)*t^(3/2)

replace t = x^4 + x

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

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

We'll use substitution to evaluate the indefinite integral.

Let x^4 + x = t

We'll differentiate both sides and we'll get:

(4x^3 + 1)dx = dt

Int (4x^3 + 1)*sqrt(x^4 + x)dx = Int (sqrt t)dt

Int (sqrt t)dt = Int t^(1/2) dt

Int t^(1/2) dt = t^(1/2 + 1)/(1/2 + 1) + C

Int t^(1/2) dt = 2t^(3/2)/3 + C

Int t^(1/2) dt = [2t*sqrt(t)]/3 + C

**Therefore, the indefinite integral of the given function is Int (4x^3 + 1)*sqrt(x^4 + x)dx = [2*(x^4 + x)*sqrt(x^4 + x)]/3 + C.**