How Can You Integrate x/(a^2+x^2)^(3/2) Without Explicit Substitution?

[abdo799]

in this video , the prof had to integrate x/(a^2+x^2)^3/2 , i know we usually do this using substitution , but in the video...he ignored the x and integrate like it was 1/(a^2+x^2)^3/2, how does that work?

 

[micromass]

You will probably need to give more information about what he did.

 

[abdo799]

he said integration of x/(a^2+x^2)^3/2= x*(-2)/(a^2+x^2)^1/2*2x
i really don't know what he did, he differentiated the bottom part then divided by new power and multiplied by differentiation of x^2

 

[abdo799]

anyway, here's the link if you want to see it , u will have to watch the first 1:30 mins only
https://courses.edx.org/courses/Ric...46b30460106/5aa2ad1721994333a179828f5660091b/

 

[abdo799]

there was a mistake with the powers in the question and i corrected it

 

[SteamKing]

Look carefully at the integrand x/(a^2+x^2)^(2/3). What is the derivative of (a^2+x^2)? Is it x times some constant perhaps? Can you rewrite the integrand as the product of two expressions, rather than the quotient?

BTW, your video requires a login to view, so we can't see it.

 

[abdo799]

Look carefully at the integrand x/(a^2+x^2)^(2/3). What is the derivative of (a^2+x^2)? Is it x times some constant perhaps? Can you rewrite the integrand as the product of two expressions, rather than the quotient?

BTW, your video requires a login to view, so we can't see it.

the power on the brakets is 3/2

 

[SteamKing]

the power on the brakets is 3/2

The power on the brackets is immaterial. The principle remains.

 

[abdo799]

The power on the brackets is immaterial. The principle remains.

i figured out what he did, he differentiated the bottom part...and that's it

 

[HallsofIvy]

In other words he did exactly the "$u= x^2+ a^2$" substitution, just not writing it out explicitly.