Thread: Using change of variables to find indefinite integrals

These problems are destroying my mind.

\int 1/((10x-3)^2)

So far I've... \int u^2 1/udu, u=10x-3

After this is where my thoughts unravel.

Maybe...

(1/10) * (u^3)/3 = (1/10) * ((10x-3)^3)/3 ????


The next one is

\int (6x+1) * \sqrt{3x^2+x}dx where u is given and u = 3x^2+x

I realize the derivative of 3x^2+x is 6x+1 and this offers some kind of portent but I fail to see the next logical step.

Any help is appreciated.

Originally Posted by AlphaTauLambda

These problems are destroying my mind.

\int 1/((10x-3)^2)

So far I've... \int u^2 1/udu, u=10x-3

After this is where my thoughts unravel.

Maybe...

(1/10) * (u^3)/3 = (1/10) * ((10x-3)^3)/3 ????


The next one is

\int (6x+1) * \sqrt{3x^2+x}dx where u is given and u = 3x^2+x

I realize the derivative of 3x^2+x is 6x+1 and this offers some kind of portent but I fail to see the next logical step.

Any help is appreciated.

Rewrite the integral and then use substitution:

I've worked out (with some help)
u =10x-3
du=10dx
1/10du=dx

1/10 * the integral of 1/u^2

I've taken this to mean

1/10 * -1/10x-3 or -1/100x-30

where -1/100x-30 is the final solution.

I'm having trouble deriving this, however. My answer comes out as 1/(100*x^2-60*x+9)

Originally Posted by AlphaTauLambda

I've worked out (with some help)
u =10x-3
du=10dx
1/10du=dx

1/10 * the integral of 1/u^2 Mr F says: Do this integral. Then substitute back u = 10x - 3.

I've taken this to mean

1/10 * -1/10x-3 or -1/100x-30 Mr F says: Yes, this is correct.

where -1/100x-30 is the final solution.

I'm having trouble deriving this, however. My answer comes out as 1/(100*x^2-60*x+9)

..

Thank you, all. You have been wonderful. I'm much more suited for chemistry and biology... The worst math I have to worry about is michaelis mentin kinematics and mathemeticians such as yourselves have already mapped that bit all out.

Thank you, again!