Question

How do you integrate `(x^3)/(1+x^2)`?

Answer 1

`x^2/2-1/2ln(x^2+1)+C.`

Explanation:

First note that the given integrand (function to be integrated) is an Improper Rational Function [degree of poly. in Numerator (Nr.) `=3>2=`degree of poly. in Deno.(Dr.)]. So, before integrating, we have to make it Proper. Usually, Long Division is performed for this, but we proceed as under:

`Nr.= x^3= x^3+x-x=x(x^2+1)-x.`
`:.` Integrand = Nr./Dr. = `{x(x^2+1)-x}/(x^2+1) ={x(x^2+1)}/(x^2+1)-x/(x^2+1)=x-x/(x^2+1)`

`:. intx^3/(x^2+1)dx=int[x-x/(x^2+1)]dx=intxdx-(1/2)int(2x)/(x^2+1)dx=x^2/2-1/2ln(x^2+1)+C.`

Notice that the second integral follows by using the formula ; `int(f'(x))/f(x)dx=lnf(x).`

Answer 2

`int x^3/(1+x^2) d x=1/2[x^2-ln(1+x^2)]+C`

Explanation:

`int x^3/(1+x^2) d x=?`

`(x^³)/(1+x^2)=x-x/(1+x^2)`

`"we can write the expression of "x^3/(1+x^2) " as "x-x/(1+x^2)" because both expressions are equal"`

`int x^3/(1+x^2) d x=int (x-x/(1+x^2))d x=int x d x-int x/(1+x^2) d x`

`int x^3/(1+x^2) d x=1/2 x^2-1/2 int (2x*d x)/(1+x^2)`

`1+x^2=u" ; "2x*d x=d u`

`int x^3/(1+x^2) d x=1/2 x^2-1/2 int (d u)/u`

`int x^3/(1+x^2) d x=1/2 x^2-1/2ln u+C`

`"plug "u=1+x^2;`

`int x^3/(1+x^2) d x=1/2 x^2-1/2ln(1+x^2)+C`

`int x^3/(1+x^2) d x=1/2[x^2-ln(1+x^2)]+C`