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I'm trying to learn how to calculate integrals on my own but I don't quite understand it.

I found the following exercise on the internet but I don't even know where to start. I'm really lost. Can you explain it to me at a level that is quite "for fools" like me?

Evaluate $$ \int_{0}^{1} \frac{x^3 - 1}{1 + x^2} \,dx. $$

Anne Bauval
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2 Answers2

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There are many ways to evaluate this integral. Here is one rule: when you see an integral going from $0$ to $1$ and there is a "$1+x^m$" in the denominator, you may need to use the geometric series: $$ \sum_{n=0}^{\infty}(-x^m)^n = \frac{1}{1+x^m}. $$ So your integral becomes $$ \int_0^1 (x^3-1)\sum_{n=0}^{\infty}(-1)^nx^{2n}dx = \sum_{n=0}^{\infty}(-1)^n(\frac{1}{2n+4} - \frac{1}{2n+1}). $$ You should recognize this sum as a combination of the series for the arctangent and the series for the natural logarithm.

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Use polynomial long division to get the integrand into the form $$P(x)+\frac{Ax+B}{1+x^2}$$ for some polynomial $P$ and constants $A,B$. Now you can integrate $\dfrac{Ax}{1+x^2}$ with the substitution $u=x^2$; and $\dfrac{B}{1+x^2}$ is well known.

TonyK
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