integration of xtan- 1x
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This can be done by integration by parts.
![I = \int\limits^a_b {x\ tan^{-1}x} \, dx \\\\Let\ u=tan^{-1}x,\ \ \ u'=\frac{1}{1+x^2},\ \ \ \ v'=x,\ \ \ v=\frac{1}{2}x^2\\\\I=u\ v- \int\limits^a_b {u'\ v} \, dx \\\\=\frac{1}{2}x^2tan^{-1}x- \frac{1}{2}\int\limits^a_b {\frac{x^2}{1+x^2}} \, dx \\\\=\frac{1}{2}x^2tan^{-1}x- \frac{1}{2}\int\limits^a_b (1-{\frac{1}{1+x^2}}) \, dx \\\\=\frac{1}{2}x^2tan^{-1}x- \frac{1}{2}x+\frac{1}{2}tan^{-1}x+K\\\\=\frac{1}{2}[(1+x^2)tan^{-1}x-x]+K](https://tex.z-dn.net/?f=I+%3D++%5Cint%5Climits%5Ea_b+%7Bx%5C+tan%5E%7B-1%7Dx%7D+%5C%2C+dx+%5C%5C%5C%5CLet%5C+u%3Dtan%5E%7B-1%7Dx%2C%5C+%5C+%5C+u%27%3D%5Cfrac%7B1%7D%7B1%2Bx%5E2%7D%2C%5C+%5C+%5C+%5C+v%27%3Dx%2C%5C+%5C+%5C+v%3D%5Cfrac%7B1%7D%7B2%7Dx%5E2%5C%5C%5C%5CI%3Du%5C+v-+%5Cint%5Climits%5Ea_b+%7Bu%27%5C+v%7D+%5C%2C+dx+%5C%5C%5C%5C%3D%5Cfrac%7B1%7D%7B2%7Dx%5E2tan%5E%7B-1%7Dx-+%5Cfrac%7B1%7D%7B2%7D%5Cint%5Climits%5Ea_b+%7B%5Cfrac%7Bx%5E2%7D%7B1%2Bx%5E2%7D%7D+%5C%2C+dx+%5C%5C%5C%5C%3D%5Cfrac%7B1%7D%7B2%7Dx%5E2tan%5E%7B-1%7Dx-+%5Cfrac%7B1%7D%7B2%7D%5Cint%5Climits%5Ea_b+%281-%7B%5Cfrac%7B1%7D%7B1%2Bx%5E2%7D%7D%29+%5C%2C+dx+%5C%5C%5C%5C%3D%5Cfrac%7B1%7D%7B2%7Dx%5E2tan%5E%7B-1%7Dx-+%5Cfrac%7B1%7D%7B2%7Dx%2B%5Cfrac%7B1%7D%7B2%7Dtan%5E%7B-1%7Dx%2BK%5C%5C%5C%5C%3D%5Cfrac%7B1%7D%7B2%7D%5B%281%2Bx%5E2%29tan%5E%7B-1%7Dx-x%5D%2BK "I = \int\limits^a_b {x\ tan^{-1}x} \, dx \\\\Let\ u=tan^{-1}x,\ \ \ u'=\frac{1}{1+x^2},\ \ \ \ v'=x,\ \ \ v=\frac{1}{2}x^2\\\\I=u\ v- \int\limits^a_b {u'\ v} \, dx \\\\=\frac{1}{2}x^2tan^{-1}x- \frac{1}{2}\int\limits^a_b {\frac{x^2}{1+x^2}} \, dx \\\\=\frac{1}{2}x^2tan^{-1}x- \frac{1}{2}\int\limits^a_b (1-{\frac{1}{1+x^2}}) \, dx \\\\=\frac{1}{2}x^2tan^{-1}x- \frac{1}{2}x+\frac{1}{2}tan^{-1}x+K\\\\=\frac{1}{2}[(1+x^2)tan^{-1}x-x]+K")
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