Integration By Parts: Solving Find \int (2x^4\ln3x)\ dx

[trollcast]

Homework Statement

Find $\int (2x^4\ln3x)\ dx$

Homework Equations

The Attempt at a Solution

I let u = ln3x and dv/dx = x⁴ and I've managed to solve it and get an answer of:

$x^5(\frac{2}{5}\ln(3x) - \frac{1}{45} + c)$

Which is really close to the answer from wolfram alpha but the last term should be 2/25 not 1/45?

 

[jedishrfu]

Can you show your intermediate steps?

What did you get for du?

what did you get for v?

and then u*v - integral(vdu) ?

 

[SammyS]

Homework Statement

Find $\int (2x^4\ln3x)\ dx$

Homework Equations

The Attempt at a Solution

I let u = ln3x and dv/dx = x⁴ and I've managed to solve it and get an answer of:

$x^5(\frac{2}{5}\ln(3x) - \frac{1}{45} + c)$

Which is really close to the answer from wolfram alpha but the last term should be 2/25 not 1/45?

How did you get the 1/45 ?

The constant of integration should outside of the parentheses .

$\displaystyle \ x^5\left(\frac{2}{5}\ln(3x) - \frac{1}{45}\right)+C\$

 

[trollcast]

Can you show your intermediate steps?

What did you get for du?

what did you get for v?

and then u*v - integral(vdu) ?

$\frac{du}{dx} = \frac{1}{3x}$
$v = \frac{x^5}{5}$
$uv - \int v\frac{du}{dx} =2( \ln3x * \frac{x^5}{5} - \int(\frac{x^5}{5} * \frac{1}{3x}))$

 

[Infrared]

What is the derivative of ln(3x)?

Edit: You can either use the chain rule as SammyS pointed out below, or you can use the product property of logs to write ln(3x)=ln3+lnx

 

[SammyS]

$\frac{du}{dx} = \frac{1}{3x}$
$v = \frac{x^5}{5}$
$uv - \int v\frac{du}{dx} =2( \ln3x * \frac{x^5}{5} - \int(\frac{x^5}{5} * \frac{1}{3x}))$

Using the chain rule:

$\displaystyle \frac{d}{dx}\, \ln(3x)=\frac{3}{3x}$

 

[trollcast]

Using the chain rule:

$\displaystyle \frac{d}{dx}\, \ln(3x)=\frac{3}{3x}$

Thanks got it now.

 

[SammyS]

What is the derivative of ln(3x)?

Edit: You can either use the chain rule as SammyS pointed out below, or you can use the product property of logs to write ln(3x)=ln3+lnx

Yes, ln3+lnx makes it even more obvious.