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