# Thread: Quick integration question

1. ## Quick integration question

Integrate wrt x, $\displaystyle \frac{1}{4x^2 + 4x +17}dx$.

I rearranged it into the form $\displaystyle \frac{1}{(2x+1)^2 + 16}dx$.

I know this integrates to some form of arctan, and I though that it would be

$\displaystyle \frac{1}{4} arctan{(\frac{2x+1}{4})}$.

But in the book they have the answer as $\displaystyle \frac{1}{8} arctan{(\frac{2x+1}{4})}$.

Is this because you need to divide by the derivative of $\displaystyle (2x+1)$ as well, or is there something I'm missing?

Thanks

2. Oh and I know I'm missing the constant, but this is part of a definite integral question.

3. Let y = 2x+1 then dy = 2 dx

$\displaystyle \int \frac{dx}{(2x+1)^2+16} = \frac12 \:\int \frac{dy}{y^2+16}$

Let y = 4z then dy = 4 dz

$\displaystyle \int \frac{dx}{(2x+1)^2+16} = \frac12 \: \int \frac{4 dz}{16(z^2+1)} = \frac18 \int \frac{dz}{z^2+1}$

$\displaystyle \int \frac{dx}{(2x+1)^2+16} = \frac18 \:\arctan(z) + c = \frac18 \:\arctan\left(\frac{2x+1}{4}\right) + c$

4. Hello,
Originally Posted by craig
Integrate wrt x, $\displaystyle \frac{1}{4x^2 + 4x +17}dx$.

I rearranged it into the form $\displaystyle \frac{1}{(2x+1)^2 + 16}dx$.

I know this integrates to some form of arctan, and I though that it would be

$\displaystyle \frac{1}{4} arctan{(\frac{2x+1}{4})}$.

But in the book they have the answer as $\displaystyle \frac{1}{8} arctan{(\frac{2x+1}{4})}$.

Is this because you need to divide by the derivative of $\displaystyle (2x+1)$ as well, or is there something I'm missing?

Thanks
It would be 1/4 if it were $\displaystyle \frac{1}{(x+\dots)^2+16}$
But here, it's as if you omit the chain rule, since there's a coefficient in front of x.

5. Thanks for both replies.

Originally Posted by Moo
It would be 1/4 if it were $\displaystyle \frac{1}{(x+\dots)^2+16}$
But here, it's as if you omit the chain rule, since there's a coefficient in front of x.
So if for example I had $\displaystyle \int \frac{1}{(3x+1)^2 + 16}dx$, the answer would be $\displaystyle \frac{1}{12} \:\arctan\left(\frac{3x+1}{4}\right) + c$

Thanks again

6. Originally Posted by craig
Thanks for both replies.

So if for example I had $\displaystyle \int \frac{1}{(3x+1)^2 + 16}dx$, the answer would be $\displaystyle \frac{1}{12} \:\arctan\left(\frac{3x+1}{4}\right) + c$

Thanks again
Yes. To see that more explicitely, use the substitution u= 3x+1.

7. Originally Posted by HallsofIvy
Yes. To see that more explicitely, use the substitution u= 3x+1.
Thanks