$\displaystyle \int{\left(\frac{x}{x^2 + 1}\right)\,dx}$
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Originally Posted by nameck $\displaystyle \int{\left(\frac{x}{x^2 + 1}\right)\,dx}$ I responded to your other post.
$\displaystyle \int{\left(\frac{x}{x^2 + 1}\right)\,dx} $ $\displaystyle u=x^2 + 1 $ $\displaystyle du=2xdx$ $\displaystyle = 0.5\int{\left(\frac{1}{u}\right)\,du} $ $\displaystyle = 0.5(Ln|x^2+1|) + C $
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