arctan integral

Alright so, I'm trying to take the integral of 3/2 / (u^2 + 3/4)

according to the book the answer is sqrt(3) * arctan(2u/sqrt(3))
but when I try to get the integral I get: 2*arctan(2u/sqrt(3)) -- I don't see any way to take a sqrt(3) out of the function.

The graphs look about the same except they are shifted vertically apart. Is this just a constant issue (are the answers basically equivilent)? Or did I mess up my algebra?

1
$$\frac{3/2}{u^2+3/4}$$

2
$$3/2 * \frac{1}{(4u^2+3)/4}$$

3
$$3/2*4 * \frac{1}{4u^2+3}$$

4
$$6* \frac{1}{3(4/3u^2+1)}$$
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 Quote by jesuslovesu Alright so, I'm trying to take the integral of 3/2 / (u^2 + 3/4) according to the book the answer is sqrt(3) * arctan(2u/sqrt(3)) but when I try to get the integral I get: 2*arctan(2u/sqrt(3)) -- I don't see any way to take a sqrt(3) out of the function. The graphs look about the same except they are shifted vertically apart. Is this just a constant issue (are the answers basically equivilent)? Or did I mess up my algebra? 1 $$\frac{3/2}{u^2+3/4}$$ 2 $$3/2 * \frac{1}{(4u^2+3)/4}$$ 3 $$3/2*4 * \frac{1}{4u^2+3}$$ 4 $$6* \frac{1}{3(4/3u^2+1)}$$

Careful. You know that

$$\int{ \frac{1}{v^2 + 1} dv = arctan(v) + c.$$

So, for your question, what is $v$?
 something was wrong in step 4. becuse $$\int { \frac{1} {px^2+1} dx = \frac{1} {\sqrt{p}} arctan(\sqrt{p} x) +c$$ here p > 0. thus,the correct answer is $$\sqrt{3} arctan(\frac{2u} {\sqrt{3}}) +c$$

Recognitions:
Homework Help

arctan integral

 Quote by jesuslovesu Alright so, I'm trying to take the integral of 3/2 / (u^2 + 3/4) according to the book the answer is sqrt(3) * arctan(2u/sqrt(3)) but when I try to get the integral I get: 2*arctan(2u/sqrt(3)) -- I don't see any way to take a sqrt(3) out of the function. The graphs look about the same except they are shifted vertically apart. Is this just a constant issue (are the answers basically equivilent)? Or did I mess up my algebra? 1 $$\frac{3/2}{u^2+3/4}$$ 2 $$3/2 * \frac{1}{(4u^2+3)/4}$$ 3 $$3/2*4 * \frac{1}{4u^2+3}$$ 4 $$6* \frac{1}{3(4/3u^2+1)}$$
Fine still step 4. :)
You know that:
$$\int \frac{du}{u ^ 2 + 1} = \arctan (u) + C$$, right?
And now, you've gotten to:
$$\int \frac{6 dx}{3 \left( \frac{4}{3} u ^ 2 + 1 \right)} = \int \frac{2 dx}{\left( \left( \frac{2}{\sqrt{3}} u \right) ^ 2 + 1 \right)}$$
Now, to find the anti-derivative of this expression, you can use the substitution: $$v = \frac{2}{\sqrt{3}} u$$.
Can you go from here? :)
 OK,let`s see it $$\int \frac{6 du}{3 \left( \frac{4}{3} u ^ 2 + 1 \right)} = \int \frac{2 du}{\left( \left( \frac{2}{\sqrt{3}} u \right) ^ 2 + 1 \right)}$$ and now we use the substitution: $$p = \frac{2}{\sqrt{3}} u$$ then we get this equation: $$u =\frac{\sqrt{3}}{2} p$$ so: $$\int \frac{2 du}{\left( \left( \frac{2}{\sqrt{3}} u \right) ^ 2 + 1 \right)}=\int \frac{2}{p^2+1}d(\frac{\sqrt{3}}{2} p) = \sqrt{3}actan(p) + c$$ finally we get the answer: $$\sqrt{3} arctan(\frac{2u} {\sqrt{3}}) +c$$
 Recognitions: Homework Help Yup, it's correct, congratulations. :)

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