Integration by Parts substitution

[revolve]

Homework Statement

<br /> \int\arctan(4t)dt<br />

Homework Equations

The Attempt at a Solution

<br /> \int\arctan(4t)dt = t\arctan(4t) -4 \int \frac{t}{1+16t^2}dt<br />

I'm stuck at this point. I think I need to make a substitution for the denominator, but I'm not sure how to go about doing so.

 

[rock.freak667]

\int \frac{t}{1+16t^2}dt


f we let u=1+16t² what is du equal to?

 

[MarcMTL]

Homework Statement

<br /> \int\arctan(4t)dt<br />

Homework Equations

The Attempt at a Solution

<br /> \int\arctan(4t)dt = t\arctan(4t) -4 \int \frac{t}{1+16t^2}dt<br />

I'm stuck at this point. I think I need to make a substitution for the denominator, but I'm not sure how to go about doing so.

It's good so far.

You can substitute for V = {1+16t^2}, so dV = 32tdt.

Then, you'll have \frac{1}{32} \int \frac{1}{V}dV.

That equates to \frac {ln(|V|)}{32}.

Multiply by -4, to obtain \frac {-ln(|V|)}{8}.

Replace V and you're done.

Good luck,

Marc.

 

[revolve]

Got it. Thank you both! Very helpful.