Интеграл x^4arctan(x)/1+x^2

[Гост]

Здравейте,може ли да ми помогнете със следния интеграл
[tex]\int (x^4arctan(x))/(1+x^2)[/tex]

[Nathi123]

I=[tex]\int \frac{ x^{4}arctgxdx }{1+ x^{2} } = \int ( x^{2}-1)arctgxdx + \int \frac{arctgxdx}{1+ x^{2} }= \int x^{2}arctgxdx- \int arctgxdx+ \frac{1}{2}arctg ^{2}x[/tex]
I=[tex]\frac{ x^{3}arctgx }{3}- \frac{1}{3} \int \frac{3 x^{2} dx}{1+ x^{2} }-xarctgx+ \int \frac{xdx}{1+ x^{2} } + \frac{1}{2}arctg ^{2}x[/tex]
I=[tex]\frac{ x^{3}arctgx }{3}-x+\int \frac{dx}{1+ x^{2} } -xarctgx+ \frac{1}{2}ln(1+ x^{2} ) + \frac{1}{2}arctg ^{2}x[/tex]
I=[tex]\frac{ x^{3}arctgx }{3}-x+(1-x)arctgx+ \frac{1}{2}ln(1+ x^{2} ) + \frac{1}{2}arctg ^{2}x+C[/tex].
I=[tex]\frac{1}{2}arctg ^{2}x+\frac{ x^{3} -3x+3}{3}arctgx+\frac{1}{2}ln(1+ x^{2} ) -x+C[/tex].