How do you find the indefinite integral of $int (x^4+x-4)/(x^2+2)$ ?

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Answer 1 · 2017-02-19T14:23:14

The answer is $=x^3/3-2x+1/2ln(x^2+2)+C$

Let's perform a long division

$color(white)(aaaa)$ $x^4$ $color(white)(aaaaaa)$ $x-4$ $color(white)(aaaa)$ $|$ $x^2+2$

$color(white)(aaaa)$ $x^4+2x^2$ $color(white)(aaaa)$ #color(white)(aaaaa)|# $x^2-2$

$color(white)(aaaaa)$ $0+2x^2+x$ $color(white)(aa)$ $-4$

$color(white)(aaaaaaaa)$ $+0+x$ $color(white)(aaaa)$ $0$

Therefore,

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

So,

$int((x^4+x-4)dx)/(x^2+2)=intx^2dx-int2dx+int(xdx)/(x^2+2)$

$=x^3/3-2x+int(xdx)/(x^2+2)$

For the last integral,

Let $u=x^2+2$ , $=>$ , $du=2xdx$

$int(xdx)/(x^2+2)=1/2int(du)/u$

$=1/2lnu$

$=1/2ln(x^2+2)$

Putting it all together

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

How do you find the indefinite integral of int (x^4+x-4)/(x^2+2)int (x^4+x-4)/(x^2+2)?