$\displaystyle \int \frac{1-x^2}{1+x^2}.dx$ Please help me, I couldnt solve it. Thank you
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Originally Posted by whitepenguin $\displaystyle \int \frac{1-x^2}{1+x^2}.dx$ Please help me, I couldnt solve it. Thank you Note that $\displaystyle \frac{1-x^2}{1+x^2} = -1 + \frac{2}{x^2 + 1}$.
Originally Posted by whitepenguin $\displaystyle \int \frac{1-x^2}{1+x^2}.dx$ Please help me, I couldnt solve it. Thank you Break it down $\displaystyle =\int\frac{1}{1+x^2}+\int\frac{x^2}{1+x^2}$. For the first, look for the $\displaystyle \arctan{x}$ For the second, divide...
Originally Posted by VonNemo19 Break it down $\displaystyle =\int\frac{1}{1+x^2}+\int\frac{x^2}{1+x^2}$. For the first, look for the $\displaystyle \arctan{x}$ For the second, divide... Actually it's $\displaystyle \int{\frac{1}{1 + x^2}\,dx} - \int{\frac{x^2}{1 + x^2}\,dx}$.
, How come I coulnt think of that. It's really simple Thank you
Originally Posted by Prove It Actually it's $\displaystyle \int{\frac{1}{1 + x^2}\,dx} - \int{\frac{x^2}{1 + x^2}\,dx}$. What am I...like -3 now?
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