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.
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.
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