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

Question

Evelyn Agard · Accepted Answer

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#

Charles Arocha · Answer

undefined To evaluate the integral ∫(x^4 + 1)/(x^2 + 1), you can perform polynomial long division or use the method of partial fractions. After performing the division, the integral simplifies to:

∫(x^4 + 1)/(x^2 + 1) dx = ∫(x^2 - x + 1) + (x - 1 + 1/(x^2 + 1)) dx

This simplifies to:

∫(x^2 - x + 1) dx + ∫(x - 1) dx + ∫(1/(x^2 + 1)) dx

From there, you can integrate each term separately. The first two terms can be integrated using the power rule for integration, and the last term can be integrated using trigonometric substitution or a direct trigonometric integral method. Once each term is integrated, you can combine them to find the final result of the integral.