INTEGRATION

[Guest]

(e^2lnx + e^2x) dx
how would you start with this problem??

 

[Unco]

G'day, Alena.

Parentheses are our friends and should not be feared.

Do you mean:
\(\displaystyle \L \int e^{(2\ln{x})} + e^{2x} \, dx\)

If so, note that
\(\displaystyle \L e^{(2\ln{x})} = (e^{\ln{x}})^2 = x^2\)

Can you integrate from there?

 

[Guest]

ok so the derivative equals:

x^2 + 2e^2x ???

 

[Unco]

I thought we were integrating?

But the derivative of \(\displaystyle x^2\) is \(\displaystyle 2x\).

Type 2e^(2x) <- note the strategically-placed parentheses.

 

[Guest]

ya my bad... when you integrate the e^2x what do you get??

 

[Unco]

The antiderivative of e^(2x) is (1/2)e^(2x) (+ C)

Check: differentiate (1/2)e^(2x).

 

[Guest]

i dont undersatnd how you get that antiderivative... can you explain??

 

[galactus]

I assume you mean:

\(\displaystyle \int{e^{2}ln(x)+e^{2x}}\)

If you mean \(\displaystyle e^{2lnx}\), then this reduces to \(\displaystyle x^{2}\)

Take the constant outside the sign of integration:

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

You can use integration by parts on the first one:

Let \(\displaystyle u=lnx\) and \(\displaystyle dv=dx\) and \(\displaystyle du=\frac{1}{x}dx\)

\(\displaystyle v=\int{dx}=x\)

\(\displaystyle \int{lnx}dx=\int{u}dv=uv-\int{v}du=xlnx-x(\frac{1}{x})dx\)

\(\displaystyle =xlnx-\int{dx}=xlnx-x+C\)

Don't forget to multiply by \(\displaystyle e^{2}\)

Use u-substitution for \(\displaystyle e^{2x}\).

Let \(\displaystyle u=2x\) and \(\displaystyle du=2dx\), \(\displaystyle \frac{1}{2}du=dx\)

\(\displaystyle \int{e^{u}}du=e^{u}\)

Sub back in:

\(\displaystyle \frac{e^{2x}}{2}\)

Add the C if you want to be particular.

 

[Guest]

can you explain how you got this??

The antiderivative of e^(2x) is (1/2)e^(2x) (+ C)

 

[galactus]

See my previous post. It's just u-substitution.

 

[Guest]

GOT IT
THANK YOU!!!!