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What is $\int \frac{4t^2}{5t-5}dt$?

I understand I can take out the constant and divide the polynomials to simplify the function. But if I didn't do the division, how then would I integrate this?

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$t^2 = (t-1)(t+1) + 1$, so $$\int\frac{4t^2}{5t-5}\,dt = \frac{4}{5}\int\frac{t^2}{t-1}\,dx = \frac{4}{5}\int\frac{(t-1)(t+1)+1}{t-1}\,dt = \frac{4}{5}\left(\int (t+1)\,dt + \int\frac{1}{t-1}\,dt\right).$$ – Arturo Magidin Jan 10 '12 at 20:14
I don't understand the question. Why wouldn't you "do the division"? – Mark McClure Jan 10 '12 at 20:14
If you "didn't do the division", then you aren't doing things right, just as if you "don't do the multiplication" you cannot figure out what $3\times(5+2)$ is equal to. – Arturo Magidin Jan 10 '12 at 20:17

Another approach (in the end, the same thing): substitute $u=5t-5$.

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Or $u=t-1$ (fewer fractions, I often make mistakes with fractions). – André Nicolas Jan 10 '12 at 20:17
The substitution $t=e^x+1$ also works nicely, but it is basically the same thing... – N. S. Jan 10 '12 at 21:56

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