1. Feb 3, 2009

### banfill_89

1. The problem statement, all variables and given/known data

$$\int$$ arctan4t dt

2. Relevant equations

3. The attempt at a solution

ive been attempting this all day. the answer is t arctan4t-1/8 ln(1+16t^2)+C

i get this asnwer up until the 1/8.......

a few steps befor emy answer i have t arctan4t - $$\int$$ 4t/ 1+16t^2

2. Feb 3, 2009

### Staff: Mentor

In your last integral, there's a fairly obvious substitution.

BTW, you need more parentheses: the integrand can be written more understandably as 4t/(1 + 16t^2)

3. Feb 3, 2009

### banfill_89

what would be the substituition? in the book after thsi they factor out a 1/8......im lost

4. Feb 3, 2009

### Dick

Substitute u=1+16t^2.

5. Feb 3, 2009

### banfill_89

??? so am i integrating by parts again?

6. Feb 3, 2009

### Dick

No, it's a u substitution.

7. Feb 3, 2009