ln as Answer in Integration Problem

[Jason76]

\(\displaystyle \int( x^{2} + \dfrac{1}{x} + e^{x} )dx\)

\(\displaystyle \int(x^{2} + x^{-1} + e^{x})dx\)

\(\displaystyle \rightarrow \dfrac{x^{3}}{3} + \dfrac{x^{0}}{0} + e^{x} + C\)

\(\displaystyle \rightarrow \dfrac{x^{3}}{3} + \ln x + e^{x} + C\)

Is this right or on the right track?

 

[Jason76]

not really.

you just say \(\displaystyle \int \frac{1}{x} \, dx=\ln (x)\)

you don't have to go through that (incorrect) business with x⁰

the rest of it is fine.

That's what I was thinking.

 

[HallsofIvy]

So you understand that \(\displaystyle \int x^n dx= \frac{1}{n+1}x^{n+ 1}+ C\) does NOT hold for n= -1?

 

[lookagain]

not really. you just say \(\displaystyle \int \frac{1}{x} \, dx=\ln (x)\)

\(\displaystyle \int \frac{1}{x} \, dx \ = \ \ln|x| \ + \ C\)


Don't forget the absolute value bars and the arbitrary constant.