Vtyjkyuk

Evaluate: $displaystyleint fracsqrt1+x^8x^13dx$

My attempt:

I have tried substituting $1+x^8=z^2$ but that did not work. I also tried writing $x^13$ in the denominator as $x^16.x^-3$ hoping that it would bring the integrand into some form but that too did not work.

The substitution $x=tan^1/4t$ gives $$intfracsqrt1+x^8x^13dx=intfrac14sin^-4tcos tdt=-frac112sin^-3t+C=-frac112bigg(fracx^81+x^8bigg)^-3/2+C.$$

Let $x^4=tan u$ so that $dx=fracsec^2 u4x^3du$

so our integral becomes

$int fracsqrt1+tan^2 ux^13.fracsec^2 u4x^3du$

or rather

$frac14int fracsec^3 utan^4 udu$

which is

$frac14int cos usin^-4 u du$

which equals

$-frac112sin^-3u +C$

now substitute back from $u$ to $x$ which I think is

$-frac112(fracx^81+x^8)^frac-32+C$

$x=dfrac1y,dx=-dfracdyy^2$

$$intdfracsqrt1+x^8x^13dx=-intdfracy^13sqrty^8+1y^4cdot y^2dy$$

Set $sqrty^8+1=u $ or $y^8+1=v$

Write the integral as $$int frac sqrt 1+frac 1x^8x^9 dx$$ Use the substitution $$1+frac 1x^8=t^2$$ to get the new integral as $$int frac -t^24dt$$ Which is very elementary to do and then after obtaining the answer resubstitute the value of $t$.

Hint We can rewrite the radical in the integrand in the familiar form $sqrt1 + u^2$ if we substitute $$u = x^4, qquad du = 4 x^3 ,dx .$$