# Thread: [SOLVED] Integration by parts, evil arctan

1. ## [SOLVED] Integration by parts, evil arctan

$
\int x~arctan(7x)~dx
$

u= arctan(7x)

du= $\frac {1}{1+49x^2}$

dv= x~dx

v= $\frac {x^2}{2}$

This is where I am at so far:
$
x^2arctan(7x) - \frac {1}{2}\int \frac {x^2}{1+49x^2}
$

2. Use polynomial long division for the second integral. You will get 2 simpler integrals.

3. Or, if you want to do things quickly, pull 49 as a factor:
$\frac{x^2}{49(\frac{1}{49} + x^2)}$

Can you see how to simplify this quickly without resorting to long division? Hint: Add $\frac{1}{49}$ and subtract $\frac{1}{49}$ in the numerator.

4. We have only talked about the long division method, but I need a little help with the 2 other integrals. What will the other two integrals be?

So I see that after long division, I get 1/49, but unfortunately I don't really know what to do next.

$
\int \frac {1}{49} (1- \frac {x^2}{1+49x^2})
$

Is this correct? What is next?

5. Originally Posted by redman223
$
\int \frac {1}{49} (1- \frac {\color{red}x^2}{1+49x^2})
$
That $x^2$ shouldn't be there. It's supposed to be:

$\frac{1}{1+49x^2}$

Once you figure that out, notice that:

$\frac{1}{1+49x^2} = \frac{1}{1+(7x)^2}$

6. $
\int \frac {1}{49} (1- \frac {1}{1+49x^2})
$

So what happens to the first part of the equation? With regards to the 1/49 and the 1-?

7. $\int \frac {1}{49} (1- \frac {1}{1+49x^2})$ can be split into:

$= \int \frac {1}{49} - \frac{1}{49} \int \frac {1}{1+49x^2}$

$= \int \frac {1}{49} - \frac{1}{49} \int \frac {1}{1+(7x)^2}$

8. it's $\int \frac 1 {49}\left({1 - \frac {1}{1 + 49x}}\right) dx$ okay? So you can split it into two:

$\int \frac 1 {49}\left({1 - \frac {1}{1 + 49x}}\right) dx = \int \frac 1 {49}\left({1}\right) dx - \int \frac 1 {49}\left({\frac {1}{1 + 49x}}\right) dx$

9. Thanks for clarifying that up for me. Now I can see a solution using u substitution.

10. Well I thought I did, but I guess not. I am stuck once again.

$
-\frac {1}{2}\int \frac {1}{49}~dx = -\frac {1}{2}(\frac {1}{49}x)
$

If thats the case, then
$
-\frac {1}{2}(\frac {1}{49}x - \frac {1}{49}\int \frac {1}{1+49x^2}~dx)
$

u= 7x
du= 7

So in order to get the 7 in there I have to multiply 1/49 by 1/7 in order to get 1/343 right? Then would the answer just be:

$
-\frac {1}{2}(\frac {arctan(7x)}{343})
$

for the last part of the equation?

11. I just noticed that you forgot to multiply by the constant 7.

$u = \arctan{7x}$

$du = \frac{7}{1+49x^2} dx$

12. Isn't that why you pull a 1/7 out in front? Which means you would multiply 1/49 and 1/7 to get 1/343?

13. Originally Posted by redman223
Isn't that why you pull a 1/7 out in front? Which means you would multiply 1/49 and 1/7 to get 1/343?

14. you could also use a trig sub, for example say tan(u)=7x

15. $= \frac{{x^2 }}
{2}\tan ^{ - 1} \left( {7x} \right) - \frac{1}
{2}\int {\frac{{x^2 }}
{{1 + 49x^2 }}} dx \hfill \\$

$= \frac{{x^2 }}
{2}\tan ^{ - 1} \left( {7x} \right) - \frac{1}
{{2 \times 49}}\int {\frac{{49x^2 }}
{{1 + 49x^2 }}} dx \hfill \\$

$= \frac{{x^2 }}
{2}\tan ^{ - 1} \left( {7x} \right) - \frac{1}
{{98}}\int {\frac{{1 + 49x^2 - 1}}
{{1 + 49x^2 }}} dx \hfill \\$

$= \frac{{x^2 }}
{2}\tan ^{ - 1} \left( {7x} \right) - \frac{1}
{{98}}\left[ {\int {dx - } \int {\frac{1}
{{1 + 49x^2 }}} dx} \right] \hfill \\$

$= \frac{{x^2 }}
{2}\tan ^{ - 1} \left( {7x} \right) - \frac{x}
{{98}} + \frac{1}
{{98}}\int {\frac{1}
{{1 + 49x^2 }}} dx \hfill \\$

$= \frac{{x^2 }}
{2}\tan ^{ - 1} \left( {7x} \right) - \frac{x}
{{98}} + \frac{1}
{{98 \times 7}}\tan ^{ - 1} \left( {7x} \right) + c \hfill \\$

$= \frac{{x^2 }}{2}\tan ^{ - 1} \left( {7x} \right) - \frac{x}{{98}} + \frac{1}{{686}}\tan ^{ - 1} \left( {7x} \right) + c \hfill \\$

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