antiderivative of 2xe^x^2 - 2 / x (check answer please)

[twinmom]

To take the antiderivative of 2xe^x^2 - 2 / x, wouldn't I end up with 1/2x^2e^x^2 - 2 + C? I'm getting a little thrown off by the e and its exponent raised to the second power.

 

[tkhunny]

That should not throw you off. It is lesson #1 in the "u-Substitution" unit.

Try u = x^2 and you should be nearly done.

 

[twinmom]

ok, so if u=x^2, then du=2x dx right? My answer comes out to be e^x^2-2+c. Is that right?

 

[tkhunny]

No. That's \(\displaystyle \L\,\frac{2}{x}\), right? That goes to 2*ln(|x|). Ringing any bells?

 

[twinmom]

I'm sorry, but no. I know that lnx is 1/x, but with this problem I'm still not getting what you mean I guess.

To take the antiderivative of 2xe^x^2-2/x, wouldn't I keep 2 as a constant and with using substitution letting u=x^2 I end up with e^x^2-2+C?

 

[tkhunny]

You tried that already.

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

You have the first piece. Your task now is the second piece.

$\displaystyle \int{\frac{2}{x}}\,dx = 2*\int{\frac{1}{x}}\,dx$

How are those bells sounding?

Please don't say '2'.

 

[twinmom]

hopefully this is it...

e^u-2lnx=e^x^2-2lnx+C

Am I even close? be gentle....

 

[tkhunny]

1) Your notation needs work. The "=" doesn't mean "=".
2) ln(|x|)

Now we're talkin'.