Indefinte Integral

[lamaclass]

I needed help with a few indefinite integrals.

1. IN x 2[sup:1jtmps9q]x^{2[/sup:1jtmps9q]}dx

My work:

u = x[sup:1jtmps9q]2[/sup:1jtmps9q]

du = 2xdx

xdx = 1/2 du

=1/2 IN 2[sup:1jtmps9q]u[/sup:1jtmps9q]du

=1/2 2[sup:1jtmps9q]u[/sup:1jtmps9q]/ln 2 + C

After this I'm not sure what to do next.

 

[Deleted member 4993]

I needed help with a few indefinite integrals.

1. IN x 2[sup:3q5og31r]x^{2[/sup:3q5og31r]}dx

My work:

u = x[sup:3q5og31r]2[/sup:3q5og31r]

du = 2xdx

xdx = 1/2 du

=1/2 IN 2[sup:3q5og31r]u[/sup:3q5og31r]du

=1/2 2[sup:3q5og31r]u[/sup:3q5og31r]/ln 2 + C

After this I'm not sure what to do next.

What do you get, if you differentiate the following:

y = 2[sup:3q5og31r]x[/sup:3q5og31r]

.

 

[lamaclass]

Oh! Differentiating it I would get

y=2[sup:2be5dys1]x[/sup:2be5dys1]/ln 2

From there though, I'm unsure of how to solve it.

 

[galactus]

You have it.

\(\displaystyle \int x\cdot 2^{x^{2}}dx\)

Let \(\displaystyle u=x^{2}, \;\ \frac{du}{2}=xdx\)

\(\displaystyle \frac{1}{2}\int 2^{u}du=\frac{1}{2}\cdot \frac{2^{u}}{ln(2)}\)

Resub:

\(\displaystyle \frac{1}{2}\frac{2^{x^{2}}}{ln(2)}+C\)

That's it.

 

[lamaclass]

Thanks for your help galactus!

I had one more like this and wanted to see if I could have someone check to see if I did it right:

IN 4[sup:1mbwiq40]x[/sup:1mbwiq40] e[sup:1mbwiq40]x[/sup:1mbwiq40]

4[sup:1mbwiq40]x[/sup:1mbwiq40]/ln 4 e[sup:1mbwiq40]x[/sup:1mbwiq40] + C

 

[Deleted member 4993]

Thanks for your help galactus!

I had one more like this and wanted to see if I could have someone check to see if I did it right:

IN 4[sup:2suhzb48]x[/sup:2suhzb48] e[sup:2suhzb48]x[/sup:2suhzb48]

4[sup:2suhzb48]x[/sup:2suhzb48]/ln 4 e[sup:2suhzb48]x[/sup:2suhzb48] + C

You have to use integration by parts here:

\(\displaystyle \int 4^x e^x dx = 4^x e^x - \int e^x [4^x * ln(4)] dx\)

\(\displaystyle \int 4^x e^x dx + ln(4) * \int e^x 4^x dx= 4^x e^x\)

\(\displaystyle \int 4^x e^x dx = \frac{4^x e^x}{1+ln(4)} + C\)