`int_(sqrt(2)/3)^(2/3)1/(x^5sqrt(9x^2-1))dx`

Let's first evaluate the indefinite integral by integral substitution,

Let `x=1/3sec(u)`

`=>dx=1/3sec(u)tan(u)du`

`int1/(x^5sqrt(9x^2-1))dx=int(1/((1/3sec(u))^5sqrt(9(1/3sec(u))^2-1)))1/3sec(u)tan(u)du`

`=int(1/(1/243sec^5(u)sqrt(sec^2(u)-1)))1/3sec(u)tan(u)du`

Now use the identity:`sec^2(theta)=1+tan^2(theta)`

`=int(243/(3sec^5(u)sqrt(1+tan^2(u)-1)))sec(u)tan(u)du`

`=int(81sec(u)tan(u))/(sec^5(u)sqrt(tan^2(u)))du`

`=81int1/(sec^4(u))du`

`=81intcos^4(u)du`

Now let's use the identity:`cos^2(theta)=(1+cos(2theta))/2`

`=81int((1+cos(2u))/2)^2du`

`=81int(1+cos^2(2u)+2cos(2u))/4du`

(The entire section contains 288 words.)

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