If $$\int f(x)dx = \psi(x)$$, then $$\int x^5 f(x^3)dx$$, is equal to

We are given that $$\int f(x)\,dx=\psi(x)$$. This immediately tells us the fundamental relation between $$f(x)$$ and $$\psi(x)$$, namely the inverse of integration:

$$\frac{d}{dx}\bigl[\psi(x)\bigr]=f(x).$$

Now we wish to evaluate the indefinite integral

$$\int x^{5}\,f(x^{3})\,dx.$$

First, we observe that the exponent of $$x$$ in the argument of $$f$$ is $$3$$, which suggests a substitution based on $$x^{3}$$. To prepare for that, we rewrite the integrand by splitting the power $$x^{5}$$ into $$x^{3}\cdot x^{2}$$:

$$\int x^{5}\,f(x^{3})\,dx=\int \bigl(x^{3}\bigr)\,\bigl(x^{2}\bigr)\,f(x^{3})\,dx.$$

We now introduce the substitution

$$t=x^{3}\quad\text{(so that }t\text{ is the inside of }f),$$

and differentiate it to find the differential $$dt$$:

$$dt=\frac{d}{dx}\bigl[x^{3}\bigr]\,dx=3x^{2}\,dx\quad\Longrightarrow\quad x^{2}\,dx=\frac{dt}{3}.$$

Substituting $$t=x^{3}$$ and $$x^{2}\,dx=\dfrac{dt}{3}$$ into the integral gives

$$\int x^{5}\,f(x^{3})\,dx=\int\bigl(x^{3}\bigr)\,f(x^{3})\;\bigl(x^{2}\,dx\bigr) =\int t\,f(t)\;\frac{dt}{3} =\frac{1}{3}\int t\,f(t)\,dt.$$

At this stage we have reduced the original integral to the simpler integral $$\int t\,f(t)\,dt$$ in the variable $$t$$. To evaluate this new integral, we employ the integration by parts formula. Recall the standard formula:

$$\int u\,dv=u\,v-\int v\,du.$$

We make the following choices:

$$u=t\quad\Longrightarrow\quad du=dt,$$

$$dv=f(t)\,dt\quad\Longrightarrow\quad v=\psi(t)\quad\bigl(\text{since }\psi'(t)=f(t)\bigr).$$

Applying integration by parts:

$$\int t\,f(t)\,dt = t\,\psi(t)-\int \psi(t)\,dt.$$

Therefore, bringing back the factor $$\tfrac13$$ we carried along earlier, we have

$$\frac{1}{3}\int t\,f(t)\,dt =\frac{1}{3}\Bigl[t\,\psi(t)-\int \psi(t)\,dt\Bigr]+C,$$

where $$C$$ represents the constant of integration.

The last task is to return to the original variable $$x$$. Remembering that $$t=x^{3}$$, we replace every $$t$$ by $$x^{3}$$:

$$\frac{1}{3}\Bigl[t\,\psi(t)-\int \psi(t)\,dt\Bigr]+C =\frac{1}{3}\Bigl[x^{3}\,\psi(x^{3})-\int \psi(t)\,dt\Bigr]+C.$$

However, the term $$\int \psi(t)\,dt$$ is still written in the variable $$t$$. To express it completely in the variable $$x$$, we transform it back using $$t=x^{3}$$ and the differential relation $$dt=3x^{2}\,dx$$:

$$\int \psi(t)\,dt =\int \psi(x^{3})\,dt =\int \psi(x^{3})\,(3x^{2}\,dx) =3\int x^{2}\,\psi(x^{3})\,dx.$$

Substituting this result into the previous expression yields

$$\frac{1}{3}\Bigl[x^{3}\,\psi(x^{3})-3\int x^{2}\,\psi(x^{3})\,dx\Bigr]+C =\frac{1}{3}x^{3}\psi(x^{3})-\int x^{2}\psi(x^{3})\,dx+C.$$

This final expression matches exactly the form given in Option A.

Hence, the correct answer is Option A.