Indefinite integral with a term to a high power

Question

$$ \int x^2(x-1)^{100} dx $$

I don't know what substitution to apply here. If we do $u=x^2$, we get $du=2x dx$ which doesn't work here. I know that the antiderivative of $(x-1)^{100}$ is $\dfrac{1}{101}(x-1)^{101} + C$, but I have no idea how to attempt this integral with a substitution.

Answer 1

HINT:

$$x^2=(x-1+1)^2=(x-1)^2+2(x-1)+1$$

$$\implies x^2(x-1)^{100}=?$$

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Alternatively, choose $$x-1=u\iff x=u+1$$

Answer 2

$$I={\int}(x-1)^{100}x^2\,dx$$ Substitute $~~x-1=u$
Differentiating b/s $~~\mathrm{d}u=\mathrm{d}x$

$$\implies I=\int u^{100}\left(u+1\right)^2\,\mathrm{d}u$$ $$=\int u^{102}+2u^{101}+u^{100}\,\mathrm{d}u$$ $$=\int u^{102}\,\mathrm{d}u+\int u^{101}\,\mathrm{d}u+\int u^{100}\,\mathrm{d}u$$ $$=\frac{u^{103}}{103}+\frac{u^{102}}{51}+\frac{u^{101}}{101}+C$$ $$=\frac{(x-1)^{103}}{103}+\frac{(x-1)^{102}}{51}+\frac{(x-1)^{101}}{101}+C$$