help with integration by parts using a trig inverse

[drakecampbell]

Apologies in advance because I don't know how to make the integration sign on the computer.

The problem is (integral)y tan^-1 3y dy.

I started with u = tan^-1 3y and du = 3/(1+9y^2) and dv = y and v = y^2/2

The result is (1/2)y^2tan^-1 (3y) - (3/2)(integral)y^2/(1+9y^2)

I'm stuck here and don't have any idea how I should progress. Thank you in advance for any suggestions.

 

[galactus]

You seem to be on to a good start.

You still have to integrate \(\displaystyle \int\frac{y^{2}}{1+9y^{2}}dy\)

Expand:

\(\displaystyle \frac{y^{2}}{1+9y^{2}}=\frac{1}{9}-\frac{1}{9(9y^{2}+1)}\)

Leading to:

\(\displaystyle \frac{1}{9}\int dy-\frac{1}{9}\int \frac{1}{9y^{2}+1}dy\)

See the rightmost integral?. It is in an arctan form. Remember?. \(\displaystyle \int\frac{1}{1+u^{2}}du=tan^{-1}(u)\)

\(\displaystyle \int \frac{1}{9y^{2}+1}dy=\int\frac{1}{1+(3y)^{2}}dy\)

Can you finish now?. What happens if you let u=3y?.