1. ## integration by parts

I am working on a integration by parts problem.

$\displaystyle \int tan^{-1}(4t)dt$

Setting $\displaystyle u=tan^{-1}(4t)$, then $\displaystyle dv=dt$, $\displaystyle dv=\frac{1}{1+(4t)^2}$ and $\displaystyle v=t$

Plugging into the equation, we get:

$\displaystyle tan^{-1}(4t)(t)-\int \frac{4t}{1+(4t)^2}dt$

I am good with all of that, except the 4 in the numerator of the integrand. I don't see where that comes from, since v=t. With the 4, it does lead to the correct answer.

Can someone explain?

2. ## Re: integration by parts

Originally Posted by baldysm
I am working on a integration by parts problem.

$\displaystyle \int tan^{-1}(4t)dt$
Look at this webpage.

3. ## Re: integration by parts

Originally Posted by baldysm
I am working on a integration by parts problem.

$\displaystyle \int tan^{-1}(4t)dt$

Setting $\displaystyle u=tan^{-1}(4t)$, then $\displaystyle dv=dt$, $\displaystyle dv=\frac{1}{1+(4t)^2}$ and $\displaystyle v=t$
You have a lot of typos that make this very difficult to read. You mean
$\displaystyle u= tan^{-1}(4t)$ and $\displaystyle dv= dt$. Then $\displaystyle du= \frac{4}{1+ (4t)^2} dt$ and $\displaystyle v= t$. An error that I think is NOT a typo is that you have $\displaystyle du= \frac{1}{1+ (4t)^2}$ without the "4" in the numerator. You need that because of the "chain rule". To differentiate $\displaystyle u= tan^{-1}(4t)$, let y= 4t so that $\displaystyle u= tan^{-1}(y)$. Then $\displaystyle \frac{du}{dy}= \frac{1}{1+ y^2}$ and $\displaystyle \frac{dy}{dx}= 4$. By the chain rule, $\displaystyle \frac{du}{dx}= \frac{du}{dy}\frac{dy}{dx}= \frac{1}{1+ u^2}(4)= \frac{4}{1+ 16t^2}$.

Plugging into the equation, we get:

$\displaystyle tan^{-1}(4t)(t)-\int \frac{4t}{1+(4t)^2}dt$

I am good with all of that, except the 4 in the numerator of the integrand. I don't see where that comes from, since v=t. With the 4, it does lead to the correct answer.

Can someone explain?