1. Integration by Substitution Discrepancy

Hi Everyone,

The integral of $\frac{1}{x^2}$ can be solved by applying the integration law onto $x^-2$ to get $\frac{-1}{x}$.

But if I try to solve this by Integration by Substitution, where $u=x^2$, I get the answer to be ;

$u=x^2$

$\frac{du}{dx} = 2x$

$dx = \frac{du}{2x}$

$\int\frac{1}{u} dx$

$\int\frac{1}{u} \frac{du}{2x}$

$\frac{1}{2x}\int\frac{1}{u} du$

$\frac{1}{2x} \cdot ln u$

$\frac{ln u}{2x}$

$\frac{ln x^2}{2x}$

But this is not equal to $\frac{-1}{x}$. Is it even possible to solve this $\int\frac{1}{x^2}$ by subsitution?

Sorry for bad layout, I will learn!

Thanks for Responses!

Roman

2. Originally Posted by romjke
Hi Everyone,

The integral of $\frac{1}{x^2}$ can be solved by applying the integration law onto $x^-2$ to get $\frac{-1}{x}$.

But if I try to solve this by Integration by Substitution, where $u=x^2$, I get the answer to be ;

$u=x^2$

$\frac{du}{dx} = 2x$

$dx = \frac{du}{2x}$

$\int\frac{1}{u} dx$

$\int\frac{1}{u} \frac{du}{2x}$

$\frac{1}{2x}\int\frac{1}{u} du$

$\frac{1}{2x} \cdot ln u$

$\frac{ln u}{2x}$

$\frac{ln x^2}{2x}$

But this is not equal to $\frac{-1}{x}$. Is it even possible to solve this $\int\frac{1}{x^2}$ by subsitution?

Sorry for bad layout, I will learn!

Thanks for Responses!

Roman
You cannot use substitution in this case (at least the way you are trying to do it). The du term (2x) would have to appear in the original integrand for the substitution to make any sense.

3. So this method of substitution cannot be used when $x$ is to a power higher than 1?

Cheers

4. The point is that you cannot have an "x" left when you change to integratind "du". And you cannot just take "x" out of the integral as you try to there because it is a function of u.

5. Originally Posted by romjke
Hi Everyone,

The integral of $\frac{1}{x^2}$ can be solved by applying the integration law onto $x^-2$ to get $\frac{-1}{x}$.

But if I try to solve this by Integration by Substitution, where $u=x^2$, I get the answer to be ;

$u=x^2$

$\frac{du}{dx} = 2x$

$dx = \frac{du}{2x}$

$\int\frac{1}{u} dx$

$\int\frac{1}{u} \frac{du}{2x}$

$\frac{1}{2x}\int\frac{1}{u} du$

$\frac{1}{2x} \cdot ln u$

$\frac{ln u}{2x}$

$\frac{ln x^2}{2x}$

But this is not equal to $\frac{-1}{x}$. Is it even possible to solve this $\int\frac{1}{x^2}$ by subsitution?

Sorry for bad layout, I will learn!

Thanks for Responses!

Roman

If $u=x^2$ then $x=\sqrt{u}$. Hence, $dx=\frac{du}{2\sqrt{u}}$.