Integration by parts using tabular method


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:

$$I:=\int x\arctan x\,dx\;\;\;\;\;,\;\;\;\begin{cases}u=\arctan x&,\;\;u'=\frac1{1+x^2}\\{}\\v'=x&,\;\;v=\frac12x^2\end{cases}\;\;\implies$$

$$I=\frac12x^2\arctan x-\frac12\int\frac{x^2}{1+x^2}dx=\frac12x^2\arctan x-\frac12\int\left(1-\frac1{1+x^2}\right)dx=$$

$$\frac12\left(x^2\arctan x-x+\arctan x\right)=\frac12\left((1+x^2)\arctan x-x\right)$$


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ryan ho
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ryan ho

Updated on August 01, 2022


  • ryan ho
    ryan ho over 1 year

    I'm trying to integrate the below using the tabular method. i do not understand the multiplication part of $$\frac{1}{1-x^{3}} and \frac{2}{1-x^{3}}$$

    $$\int xtanh^{-1}dx$$

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