If `int f(x)dx=psi(x)`, then `int x^5f(x^3)dx`

Options

A. ` (1)/(3)[x^(3)Psi(x^(3))-int x^(2) Psi(x^(3))dx]+c`
B. ` (1)/(3)x^(3)Psi(x^(3))-3int x^(3) Psi(x^(3))dx+c`
C. ` (1)/(3)x^(3)Psi(x^(3))-int x^(2) Psi(x^(3))dx+c`
D. ` (1)/(3)[x^(3)Psi(x^(3))-int x^(3) Psi(x^(3))dx]+c`

Explanation

To solve the integral \( I = \int x^5 f(x^3) \, dx \), we can follow these steps:

### Step 1: Substitution
Let \( t = x^3 \). Then, we differentiate to find \( dt \):
\[ dt = 3x^2 \, dx \quad \Rightarrow \quad dx = \frac{dt}{3x^2} \]
Also, from \( t = x^3 \), we have \( x = t^{1/3} \), so:
\[ x^2 = (t^{1/3})^2 = t^{2/3} \]
Substituting these into the integral gives:
\[ I = \int x^5 f(x^3) \, dx = \int (t^{1/3})^5 f(t) \cdot \frac{dt}{3t^{2/3}} = \int \frac{t^{5/3} f(t)}{3t^{2/3}} \, dt = \frac{1}{3} \int t^{5/3 - 2/3} f(t) \, dt = \frac{1}{3} \int t^{1} f(t) \, dt \]

### Step 2: Apply Integration by Parts
Now we can apply integration by parts to the integral \( \int t f(t) \, dt \). Let:
- \( u = t \) (first function)
- \( dv = f(t) \, dt \) (second function)

Then, we differentiate and integrate:
- \( du = dt \)
- \( v = \psi(t) \) (since \( \int f(t) \, dt = \psi(t) \))

Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \):
\[ \int t f(t) \, dt = t \psi(t) - \int \psi(t) \, dt \]

### Step 3: Substitute Back
Substituting back into our expression for \( I \):
\[ I = \frac{1}{3} \left( t \psi(t) - \int \psi(t) \, dt \right) \]
Now substituting \( t = x^3 \):
\[ I = \frac{1}{3} \left( x^3 \psi(x^3) - \int \psi(x^3) \, dx \right) \]

### Final Result
Thus, the final expression for the integral \( I \) is:
\[ I = \frac{1}{3} \left( x^3 \psi(x^3) - \int \psi(x^3) \, dx \right) \]