Integrate this

**mathproblem** · January 21st 2006, 02:48 PM

How would you integrate this [ 1 / (49*e^(-4x) + e^(4x)) ] im stuck.

**TD!** · January 21st 2006, 02:59 PM

$\int {\frac{1}{{49e^{ - 4x} + e^{4x} }}dx} = \int {\frac{{e^{4x} }}{{49 + e^{8x} }}dx} = \frac{1}{{49}}\int {\frac{{e^{4x} }}{{1 + \frac{{e^{8x} }}{{49}}}}dx}$

$\frac{1}{{49}}\int {\frac{{e^{4x} }}{{1 + \left( {\frac{{e^{4x} }}{7}} \right)^2 }}dx} \mathop \to \limits_{dt = 4e^{4x} dx}^{t = e^{4x} } \frac{1}{{49}}\frac{1}{4}\int {\frac{1}{{1 + \left( {\frac{t}{7}} \right)^2 }}dt}$

$\frac{7}{{4 \cdot 49}}\int {\frac{1}{{1 + \left( {\frac{t}{7}} \right)^2 }}d\left( {\frac{t}{7}} \right)} = \frac{1}{{28}}\arctan \left( {\frac{t}{7}} \right) + c$

And finally, substituting back:

$\frac{1}{{28}}\arctan \left( {\frac{{e^{4x} }}{7}} \right) + c$

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