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compute intergration:x(x^2-1)^7
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\[ \int x(x^2 - 1)^{7} dx \]Make u = x^2 - 1, then du = 2x. So we have: \[ \frac{1}{2} \int u^7 dx \]
How did you get 1/2?
It's that as du = 2xdx, we need to multiply the whole integral by 2/2. We get: \[ \frac{2}{2} \int x (x^2 - 1)^7 dx = \frac{1}{2} \int 2x(x^2 - 1)^7 dx \]Now note that we have du outside the parenthesis in the integral?
In general, if, after calling u = something, du = k(what is left), k being a constant, we have to multiply the integral by k/k, then rewrite the integral as: \[\frac{k}{k} \int f(x)dx = \frac{1}{k} \int k f(x)dx = \frac{1}{k} \int f(u)du\]
In our case, k = 2. :-)
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