Question

integrate (x)(x^2-1)dx by subst without first multiplying. Why not?

Answer 1

Just to get the practice of using a u-substitution; you COULD do it by multiplying, but that's seemingly not the point.

Answer 2

\[\int\limits_{-\infty}^{\infty}(x ^{3}-x)dx\]

Start by separating the terms along with the differential:
\[\int\limits_{-\infty}^{\infty}x ^{3}dx - \int\limits_{-\infty}^{\infty}xdx\]

Integrate both parts:
\[(\frac{1}{4}x^4-\frac{1}{2}x^2+c)|^{\infty}_{-\infty}\]

I'm posting the above because I just took all the time to make the equations look nice in the editor, but I know realize you were not supposed to multiply first. In this case, you were asked to integrate as given to practice u-substitution as stated earlier.

\[u = x^2-1 \]
\[du = 2xdx\]
\[\frac{1}{2}du=xdx\]
\[\frac{1}{2}\int\limits_{-\infty}^{\infty}udu\]
\[(\frac{1}{2})(\frac{1}{2})u^2=\frac{1}{4}u^2+c\]
Then substitute back in the value of u:
\[=\frac{1}{4}(x^2-1)+c\]