How do you evaluate the integral #int (x^4+1)/(x^2+1)#?

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Answer 1 · 2017-01-08T13:59:23

The answer is #=x^3/3-x+2arctanx+C#

Since the degree of the numerator is not less than the degree of the denominator, perform a long division

#color(white)(aaaa)##x^4##color(white)(aaaaaaaa)##+1##color(white)(aaaa)##∣##x^2+1#

#color(white)(aaaa)##x^4+x^2##color(white)(aaaaaaaa)####color(white)(aa)##∣##x^2-1#

#color(white)(aaaa)##0-x^2##color(white)(aaaa)##+1#

#color(white)(aaaaaa)##-x^2##color(white)(aaaa)##-1#

#color(white)(aaaaaa)##0##color(white)(aaaaaaaa)##2#

Therefore,

#(x^4+1)/(x^2+1)=x^2-1+2/(x^2+1)#

#int((x^4+1)dx)/(x^2+1)=intx^2dx-int1dx+2intdx/(x^2+1)#

#=x^3/3-x+2intdx/(x^2+1)#

Let #x=tan theta#, #=>#, #dx=sec^2theta d theta#

and #x^2+1=tan^2 theta+1=sec^2 theta#

Therefore,

#2intdx/(x^2+1)=2int(sec^2 theta d theta)/sec^2theta=2intd theta=2 theta=2arctanx#

So,

#int((x^4+1)dx)/(x^2+1)=x^3/3-x+2arctanx+C#