Obliczamy całkę metodą podstawiania: \[ \begin{split} \int \frac{e^{4x}-e^{-x}}{4}dx &=\int \frac{\dfrac{e^{5x}}{e^x}-\dfrac{1}{e^x}}{4}dx=\\[6pt] &=\int \frac{e^{5x}-1}{4e^x}dx=\\[6pt] &=\begin{vmatrix} t=e^x\\dt=e^xdx\\dx=\frac{dt}{e^x}\end{vmatrix}=\\[6pt] &=\int \frac{t^5-1}{4t}\cdot \frac{dt}{t}=\\[6pt] &=\frac{1}{4}\int\frac{t^5-1}{t^2}dt=\\[6pt] &=\frac{1}{4} \left ( \int t^3 dt-\int t^{-2} dt \right )=\\[6pt] &=\frac{1}{4} \left ( \frac{t^4}{4}-\frac{t^{-1}}{-1}+C \right )=\\[6pt] &=\frac{t^4}{16}+\frac{1}{4t}+C=\\[6pt] &=\frac{t^5+4}{16t}+C=\\[6pt] &=\frac{e^{5x}+4}{16e^x}+C \end{split} \]