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

Question

${1 \over 3}\left[ {{x^3}\psi \left( {{x^3}} \right) - \int {{x^2}\psi \left( {{x^3}} \right)dx} } \right] + C$

${1 \over 3}{x^3}\psi \left( {{x^3}} \right) - 3\int {{x^3}\psi \left( {{x^3}} \right)dx} + C$

${1 \over 3}{x^3}\psi \left( {{x^3}} \right) - \int {{x^2}\psi \left( {{x^3}} \right)dx} + C$

${1 \over 3}\left[ {{x^3}\psi \left( {{x^3}} \right) - \int {{x^3}\psi \left( {{x^3}} \right)dx} } \right] + C$

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