Integration by parts using tabular method
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I think the tabular method is mainly integration by parts done several steps at once, and in this particular case it doesn't really make things different as there's only one step:
$$I:=\int x\arctan x\,dx\;\;\;\;\;,\;\;\;\begin{cases}u=\arctan x&,\;\;u'=\frac1{1+x^2}\\{}\\v'=x&,\;\;v=\frac12x^2\end{cases}\;\;\implies$$
$$I=\frac12x^2\arctan x\frac12\int\frac{x^2}{1+x^2}dx=\frac12x^2\arctan x\frac12\int\left(1\frac1{1+x^2}\right)dx=$$
$$\frac12\left(x^2\arctan xx+\arctan x\right)=\frac12\left((1+x^2)\arctan xx\right)$$
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ryan ho
Updated on August 01, 2022Comments

ryan ho over 1 year
I'm trying to integrate the below using the tabular method. i do not understand the multiplication part of $$\frac{1}{1x^{3}} and \frac{2}{1x^{3}}$$
$$\int xtanh^{1}dx$$