Power Rule with ln and Integration

[Jason76]

How would the integration power rule be applied to 2lnx?

$\displaystyle e^{\int \dfrac{2}{x}}$ This has to be rewritten to perform integration

$\displaystyle e^{\int 2lnx}$ Rewritten as a ln

$\displaystyle e^{\int \dfrac{2lnx^{2}}{2}}$ Using "integration power rule" amd then canceling out denominator and coefficient.

$\displaystyle e^{lnx^{2}} = x^{2}$ Simplifies to $\displaystyle x^{2}$. Why?

 

[srmichael]

How would the integration power rule be applied to 2lnx?

$\displaystyle e^{\int \dfrac{2}{x}}$ This has to be rewritten to perform integration

$\displaystyle e^{\int 2lnx}$ Rewritten as a ln

$\displaystyle e^{\int \dfrac{2lnx^{2}}{2}}$ Using "integration power rule" amd then canceling out denominator and coefficient.

$\displaystyle e^{lnx^{2}} = x^{2}$ Simplifies to $\displaystyle x^{2}$. Why?

Careful:

$\displaystyle e^{\int \dfrac{2}{x}dx}=e^{2\ln(x)}=e^{\ln(x^2)}=x^2$

The last step is because e^x and ln(x) are inverses of each other. In fact, as a general rule: $\displaystyle b^{\log_{b}f(x)}=f(x)$

 

[lookagain]

> > Careful: < <

$\displaystyle e^{\int \dfrac{2}{x}dx}=e^{2\ln(x)}=e^{\ln(x^2)}=x^2$

Careful:


$\displaystyle {\int \dfrac{2}{x}dx}=$

$\displaystyle 2\ln|x| + C$

 

[dsk2]

Jason76,

Although the answer to your question is simply $\displaystyle e^{ln(x^2) +C}$ or $\displaystyle C_1 x^2$ (where $\displaystyle C_1=e^C$), if you had a question in the exam that involved integrating ln(x), know that the integral of ln(x) is x*ln(x) -x +C. It would help to derive and memorize this standard integral formula for ln(x).

Cheers,
Sai.

Edit: 01/07/13

 

[Jason76]

Is the answer $\displaystyle 2 \ln | x | + C$ or $\displaystyle x^{2} + C$

 

[Deleted member 4993]

Is the answer $\displaystyle 2 ln | x |C$ or $\displaystyle x^{2} + C$

The answer is:

$\displaystyle e^{\int \frac{2}{x} dx} \ =$

e ^2ln|x|+C = C₁*e ^2ln|x| = C₁* x²