How to do $\int_{ }^{ }\frac{e^{12t} }{e^{24t}+2e^{12t}+1}dt$

Question

How do I do the following:

$$\int_{ }^{ }\frac{e^{12t}}{e^{24t}+2e^{12t}+1}dt$$

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Answer 1

set $u=e^{12t}$, $du=12udt$, $\int{{e^{12t}\over{e^{24t}+2e^{12t}+1}}}dt$

$={1\over{12}}\int{1\over{(1+u)^2}}du=-{1\over{12}}{1\over{1+u}}=-{1\over{12}}{1\over{1+e^{12t}}}.$

Answer 2

HINT

Solve by substituition with $e^{12t}=x\implies 12e^{12t}dt=dx$

$$\int_{ }^{ }\frac{e^{12t}}{e^{24t}+2e^{12t}+1}dt=\frac1{12}\int_{ }^{ }\frac{1}{x^2+2x+1}dx=\frac1{12}\int_{ }^{ }\frac{1}{(x+1)^2}dx$$

Answer 3

Hint

$$\frac{e^{12t}}{e^{24t}+2e^{12t}+1}=\frac{e^{12t}}{(e^{12t}+1)^2}=(e^{12t}+1)^{-2}e^{12t}$$

Answer 4

Here is a method that is a little different in that it depends on a knowledge of the hyperbolic functions.

\begin{align*} \int \frac{e^{12 t}}{e^{24 t} + 2 e^{12 t} + 1} \, dt &= \int \frac{e^{12 t}}{e^{12 t} (e^{12 t} + 2 + e^{-12 t})} \, dt\\ &= \int \frac{dt}{(e^{6t} + e^{-6t})^2}\\ &= \frac{1}{4} \int \frac{dt}{\left (\dfrac{e^{6t} + e^{-6t}}{2} \right )^2}\\ &= \frac{1}{4} \int \frac{dt}{\cosh^2 (6t)}\\ &= \frac{1}{4} \int \text{sech}^2 (6t) \, dt\\ &= \frac{1}{24} \tanh (6t) + C. \end{align*}