$\displaystyle
\int x~arctan(7x)~dx
$
u= arctan(7x)
du= $\displaystyle \frac {1}{1+49x^2}$
dv= x~dx
v= $\displaystyle \frac {x^2}{2}$
This is where I am at so far:
$\displaystyle
x^2arctan(7x) - \frac {1}{2}\int \frac {x^2}{1+49x^2}
$
$\displaystyle
\int x~arctan(7x)~dx
$
u= arctan(7x)
du= $\displaystyle \frac {1}{1+49x^2}$
dv= x~dx
v= $\displaystyle \frac {x^2}{2}$
This is where I am at so far:
$\displaystyle
x^2arctan(7x) - \frac {1}{2}\int \frac {x^2}{1+49x^2}
$
Or, if you want to do things quickly, pull 49 as a factor:
$\displaystyle \frac{x^2}{49(\frac{1}{49} + x^2)}$
Can you see how to simplify this quickly without resorting to long division? Hint: Add $\displaystyle \frac{1}{49}$ and subtract $\displaystyle \frac{1}{49}$ in the numerator.
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.
$\displaystyle
\int \frac {1}{49} (1- \frac {x^2}{1+49x^2})
$
Is this correct? What is next?
it's $\displaystyle \int \frac 1 {49}\left({1 - \frac {1}{1 + 49x}}\right) dx$ okay? So you can split it into two:
$\displaystyle \int \frac 1 {49}\left({1 - \frac {1}{1 + 49x}}\right) dx = \int \frac 1 {49}\left({1}\right) dx - \int \frac 1 {49}\left({\frac {1}{1 + 49x}}\right) dx$
Well I thought I did, but I guess not. I am stuck once again.
$\displaystyle
-\frac {1}{2}\int \frac {1}{49}~dx = -\frac {1}{2}(\frac {1}{49}x)
$
If thats the case, then
$\displaystyle
-\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:
$\displaystyle
-\frac {1}{2}(\frac {arctan(7x)}{343})
$
for the last part of the equation?
$\displaystyle = \frac{{x^2 }}
{2}\tan ^{ - 1} \left( {7x} \right) - \frac{1}
{2}\int {\frac{{x^2 }}
{{1 + 49x^2 }}} dx \hfill \\$
$\displaystyle = \frac{{x^2 }}
{2}\tan ^{ - 1} \left( {7x} \right) - \frac{1}
{{2 \times 49}}\int {\frac{{49x^2 }}
{{1 + 49x^2 }}} dx \hfill \\$
$\displaystyle = \frac{{x^2 }}
{2}\tan ^{ - 1} \left( {7x} \right) - \frac{1}
{{98}}\int {\frac{{1 + 49x^2 - 1}}
{{1 + 49x^2 }}} dx \hfill \\$
$\displaystyle = \frac{{x^2 }}
{2}\tan ^{ - 1} \left( {7x} \right) - \frac{1}
{{98}}\left[ {\int {dx - } \int {\frac{1}
{{1 + 49x^2 }}} dx} \right] \hfill \\$
$\displaystyle = \frac{{x^2 }}
{2}\tan ^{ - 1} \left( {7x} \right) - \frac{x}
{{98}} + \frac{1}
{{98}}\int {\frac{1}
{{1 + 49x^2 }}} dx \hfill \\$
$\displaystyle = \frac{{x^2 }}
{2}\tan ^{ - 1} \left( {7x} \right) - \frac{x}
{{98}} + \frac{1}
{{98 \times 7}}\tan ^{ - 1} \left( {7x} \right) + c \hfill \\$
$\displaystyle = \frac{{x^2 }}{2}\tan ^{ - 1} \left( {7x} \right) - \frac{x}{{98}} + \frac{1}{{686}}\tan ^{ - 1} \left( {7x} \right) + c \hfill \\ $