# Integration by Parts substitution

## Homework Statement

$$\int\arctan(4t)dt$$

## The Attempt at a Solution

$$\int\arctan(4t)dt = t\arctan(4t) -4 \int \frac{t}{1+16t^2}dt$$

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

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

## Homework Statement

$$\int\arctan(4t)dt$$

## The Attempt at a Solution

$$\int\arctan(4t)dt = t\arctan(4t) -4 \int \frac{t}{1+16t^2}dt$$

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.

Got it. Thank you both! Very helpful.