Para encontrarmos a base para H realizaremos os cálculos abaixo:

\(\begin{align} & v(t)=[{{e}^{(-bt)}}]\cdot (a+bt) \\ & \\ & H(t)=\int_{{{b}_{1}}}^{{{b}_{2}}}{v(t)} \\ & H(t)=\int_{1/\surd LC}^{R/\left( 2L \right)}{\left( [{{e}^{(-bt)}}]\cdot (a+bt) \right)}db \\ & H(t)=\int_{1/\surd LC}^{R/\left( 2L \right)}{\left( a{{e}^{(-bt)}}+bt{{e}^{(-bt)}}) \right)}db \\ & H(t)=\left( \frac{1}{-t}{{e}^{(-bt)}}+\frac{bt{{e}^{(-bt)}}}{{{t}^{2}}}(-bt-1) \right)_{1/\sqrt{LC}}^{R/2L} \\ & H(t)=\left( \frac{1}{-R/2L}{{e}^{(-bR/2L)}}+\frac{bR/2L{{e}^{(-bt)}}}{\frac{{{R}^{2}}}{4{{L}^{2}}}}\left( \frac{-bR}{2L}-1 \right) \right)-\left( \frac{1}{-1/\sqrt{LC}}{{e}^{(b/\sqrt{LC})}}+\frac{b1/\sqrt{LC}{{e}^{(b/\sqrt{LC})}}}{{{t}^{2}}}\left( -\frac{b}{\sqrt{LC}}-1 \right) \right) \\ \end{align}\ \)

Portanto, a base para H será :

\(\begin{align} & H(t)=\left( \frac{1}{-R/2L}{{e}^{(-bR/2L)}}+\frac{bR/2L{{e}^{(-bt)}}}{\frac{{{R}^{2}}}{4{{L}^{2}}}}\left( \frac{-bR}{2L}-1 \right) \right)-\left( \frac{1}{-1/\sqrt{LC}}{{e}^{(b/\sqrt{LC})}}+\frac{b1/\sqrt{LC}{{e}^{(b/\sqrt{LC})}}}{{{t}^{2}}}\left( -\frac{b}{\sqrt{LC}}-1 \right) \right) \\ \end{align}\ \)