Integration with arctan

**Archduke01** · March 10th 2010, 02:41 PM

$\int x arctan (7x) dx$

I end up with $\int x arctan (7x) = x / (7x^2+1) - \int 1 / (7x^2+1)$

I don't know how to evaluate the integral for that last one. Am I supposed to use x as my dv instead?

**skeeter** · March 10th 2010, 03:04 PM

Originally Posted by Archduke01

$\int x arctan (7x) dx$

I end up with $\int x arctan (7x) = x / (7x^2+1) - \int 1 / (7x^2+1)$

I don't know how to evaluate the integral for that last one. Am I supposed to use x as my dv instead?

$u = \arctan(7x)$

$dv = x \, dx<br />$

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

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

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

$\int x \arctan(7x) \, dx = \frac{x^2\arctan(7x)}{2} - \frac{1}{14} \int \frac{49x^2+1-1}{1 + 49x^2} \, dx$

$\int x \arctan(7x) \, dx = \frac{x^2\arctan(7x)}{2} - \frac{1}{14} \int 1 - \frac{1}{1 + 49x^2} \, dx$

$\int x \arctan(7x) \, dx = \frac{x^2\arctan(7x)}{2} - \frac{1}{98} \int 7 - \frac{7}{1 + 49x^2} \, dx$

$\int x \arctan(7x) \, dx = \frac{x^2\arctan(7x)}{2} - \frac{x}{14} + \frac{\arctan(7x)}{98} + C$

$\int x \arctan(7x) \, dx = \left(\frac{x^2}{2}+\frac{1}{98}\right)\arctan(7x) - \frac{x}{14} + C$

**Archduke01** · March 10th 2010, 03:26 PM

^How did you manage to take out 1/14 and why did you add a negative one to the numerator?

**skeeter** · March 10th 2010, 03:35 PM

$\int \frac{1}{2} \cdot \frac{7x^2}{1 + 49x^2} \, dx$

$\frac{1}{2} \int \frac{7x^2}{1 + 49x^2} \, dx$

$\frac{1}{2} \int \frac{7}{7} \cdot \frac{7x^2}{1 + 49x^2} \, dx$

$\frac{1}{14} \int \frac{49x^2}{1 + 49x^2} \, dx$

$\frac{1}{14} \int \frac{49x^2+1-1}{1 + 49x^2} \, dx$

$\frac{1}{14} \int \frac{49x^2+1}{1 + 49x^2} - \frac{1}{1+49x^2} \, dx$

$\frac{1}{14} \int 1 - \frac{1}{1+49x^2} \, dx$

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