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

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Answer 1 · 2017-01-18T14:13:15

The answer is $=x^3/3-x+arctan(x)+C$

We need

$intx^ndx=x^(n+1)/(n+1)+C(n!=-1)$

We perform a polynomial long division

$color(white)(aaaa)$ $x^4$ $color(white)(aaaaaaaaa)$ $∣$ $x^2+1$

$color(white)(aaaa)$ $x^4+x^2$ $color(white)(aaaaa)$ $∣$ $x^2-1$

$color(white)(aaaaa)$ $0-x^2$

$color(white)(aaaaaaa)$ $-x^2-1$

$color(white)(aaaaaaaaa)$ $0+1$

Therefore,

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

So,

$int(x^4dx)/(x^2+1)=int(x^2-1)dx+color(red)(intdx/(x^2+1))$

$=x^3/3-x+color(red)(intdx/(x^2+1))$

The integral in red is done by trigonometric substitution

Let $x=tantheta$ , $=>$ , $dx=sec^2theta d theta$

and, $x^2+1=tan^2theta+1=sec^2theta$

Therefore,

$color(red)(intdx/(x^2+1)=int(sec^2theta d theta)/sec^2theta=intd theta=theta=arctan(x))$

Putting it all together

$int(x^4dx)/(x^2+1)=x^3/3-x+arctan(x)+C$

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