Question

Evaluate `int(x^6+1)/(x^2+1)dx` ?

Answer 1

`int (x^6+1)/(x^2+1) dx = x^5/5 -x^3/3 +x+C`

Explanation:

Substitute `x= tant`, `dx = sec^2tdt`:

`int (x^6+1)/(x^2+1) dx = int (tan^6t+1)/(tan^2t +1) sec^2tdt`

Use now the trigonometric identity:

`tan^2 alpha +1 = sin^2alpha/cos^2alpha +1 = (sin^2alpha +cos^2alpha)/cos^2alpha = 1/cos^2alpha = sec^2alpha`

to get:

`int (x^6+1)/(x^2+1) dx = int (tan^6t+1)/sec^2t sec^2tdt = int (tan^6t+1)dt`

Using the linearity of the integral:

`(1) " " int (x^6+1)/(x^2+1) dx = int tan^6tdt +int dt = t + int tan^6tdt`

Solve now:

`int tan^6tdt = int tan^4t tan^2t dt = int tan^4 (sec^2t-1)dt`

`int tan^6tdt = int tan^4 sec^2tdt -int tan^4dt`

`int tan^6tdt = int tan^4 d(tant) -int tan^2 (sec^2t-1)dt`

`int tan^6tdt =tan^5t/5 -int tan^2 sec^2tdt + int tan^2tdt`

`int tan^6tdt =tan^5t/5 - tan^3t/3 + int (sec^2t -1)dt`

`int tan^6tdt =tan^5t/5 - tan^3t/3 + int sec^2tdt -intdt`

`int tan^6tdt =tan^5t/5 - tan^3t/3 + tant - t +C`

Substituting in `(1)`:

`int (x^6+1)/(x^2+1) dx = t + tan^5t/5 - tan^3t/3 + tant - t +C`

`int (x^6+1)/(x^2+1) dx = tan^5t/5 - tan^3t/3 + tant + C`

and undoing the substitution:

`int (x^6+1)/(x^2+1) dx = x^5/5 -x^3/3 +x+C`

Note: from the result I realized that:
`(x^6+1)/(x^2+1) = x^4-x^2+1`
My fault not to remember the simple algebraic formula:
`(x^n+1) = (x+1)(x^(n-1)-x^(n-2) +x^(n-3) -...+1)`
valid for `n` odd where in this case `x^2` is in place of `x`.*

Answer 2

`1/5x^5-1/3x^3+x+C`

Explanation:

When integrating a rational function where the numerator has a degree equal to or higher than the denominator, start by doing "algebraic division" of the numerator by the denominator:

`\ \ \ \ \ \ \ \ \ \ \ \ ul(x^4\ \ \ \ \ \ \ \ -x^2\ \ \ \ \ \ \ \ +1\ \ )`
`x^2+1|x^6\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +1`
`\ \ \ \ \ \ \ \ \ \ \ \ ul(x^6+x^4)`
`\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -x^4`
`\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ul(-x^4-x^2)`
`\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x^2\ \ \ \ \ \ \ \ +1`
`\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ul(x^2\ \ \ \ \ \ \ \ +1)`
`\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ul(ul(0))`

In this unusual trick question the remainder is zero, so the denominator was a factor of the numerator, so the fraction is the polynomial `x^4-x^2+1` which trivially integrates to the answer given. Sneaky!

Answer 3

`x^5/5-x^3/3+x+C`

Explanation:

Consider the formula for a Geometric Series:
`a+ar+ar^2+...+ar^(n-1)=a((1-r^n)/(1-r))`
Let `a=1`, `r=-x^2`, `n=3`:
`1-x^2+x^4=1((1-(-x^2)^3)/(1-(-x^2)))`
`=(x^6+1)/(x^2+1)`
Hence `int(x^6+1)/(x^2+1)dx=int(x^4-x^2+1)dx`
`=x^5/5-x^3/3+x+C`