# Evaluate the indefinite integral of y=x^2(x^3+1)^4.

Asked on by lillyan

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

We have to find the integral of y=x^2(x^3+1)^4.

Let t = x^3+1

dt/dx = 3x^2

=> x^2 dx = (1/3) dt

Now Int [ x^2(x^3+1)^4 dx]

=> Int [ (1/3)* t^4 dt]

=> (1/3) t^5 / 5

=> t^5 / 15

replace t with x^3 + 1

=> (1/15)*(x^3 + 1)^5 + C

Therefore the required integral is (1/15)*(x^3 + 1)^5 + C.

neela | High School Teacher | (Level 3) Valedictorian

Posted on

To find the integral of x^2(x^3+1)^4.

Let f(x) = x^2(x^3+1)^4.

We put (x^3+1) = t. We differentiate (x^3+1) = t with respect to x..

3x^2 dx = dt. So x^2 dx = dt/3.

So we substitute x^3+1 = t and x^2 dx = dt/3 in  f(x) = x^2 dx and integrate.

So Int f(x) dx = Int t^4 *dt/3

Int f(x) dx = (1/3) (1/5) t^5 + C.

f(x) dx = (1/15)(x^3+1)^5. + C.

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

To evaluate the integral, we'll change the variable:

1 + x^3 = t

We'll differentiate both sides:

3x^2dx = dt

x^2dx = dt/3

We'll re-write the integral in t:

Int x^2(x^3+1)^4 dx = Int t^4 dt/3

Int t^4 dt/3 = (1/3)Int  t^4 dt

(1/3)Int  t^4 dt = (1/3)*(t^5/5) +C

(1/3)Int  t^4 dt = t^5/15 + C

Int x^2(x^3+1)^4 dx = (x^3+1)^5/15 + C

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