Integration of Exponential Functions

The basic formulas are \[ \int e^x\, dx = e^x + C, \quad \int a^x\, dx = \frac{a^x}{\ln(a)} +C.\]

Find the indefinite integral \[\int (3e^x+2^x)\, dx,\] using \(C\) as the constant of integration.

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We have \[\begin{align} \int (3e^x+2^x)\, dx &=3\int e^x dx+\int 2^x\, dx \\ &=3e^x+\frac{2^x}{\ln 2}+C, \end{align}\] where \(C\) is the constant of integration. \(_\square\)

Find the indefinite integral \[\int e^{x+2}\, dx,\] using \(C\) as the constant of integration.

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We have \[\begin{align} \int e^{x+2}\, dx &=\int e^x e^2\, dx \\ &=e^2\int e^x\, dx \\ &=e^2 e^x +C \\ &=e^{x+2}+C, \end{align}\] where \(C\) is the constant of integration. \(_\square\)

Case 1: Suppose we have an exponential function clubbed as \(\displaystyle \int e^x\big(f(x) + f'(x)\big)\, dx\). In this case, the integral is \(e^xf(x) + C.\)

Find the indefinite integral \[\int e^x\big(\sin(x) + \cos(x)\big)\, dx\] using \(C\) as the constant of integration.

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We have the integral in the form of \(\displaystyle \int e^x\big(f(x) + f'(x)\big)\, dx,\) where \(f(x) = \sin(x)\). So our integral is \[e^x\sin(x) + C.\ _\square\]

Case 2: Suppose we have an integration of the form \(\displaystyle I = \int e^{ax}\cos(bx+c).\) Its integral is \[I=\dfrac{e^{ax}\big(a\cos(bx+c)+b\sin(bx+c)\big)}{a^2 + b^2}.\]

We'll integrate the above using integration by parts as follows: \[\begin{align} I &= \int e^{ax}\cos(bx+c)\\ &= \cos(bx+c)\frac{e^{ax}}{a} + \frac{b}{a} \int e^{ax}\sin(bx+c)\\ &= \cos(bx+c)\frac{e^{ax}}{a} + \frac{b}{a}\left(\frac{e^{ax}}{a}\sin(bx+c) - \frac{b}{a} \int e^{ax}\cos(bx+c)\right)\\ &=\frac{e^{ax}\big(a\cos(bx+c)+b\sin(bx+c)\big)}{a^2} - \frac{b^2}{a^2}I. \end{align}\] So \[\begin{align} I\left(1+\frac{b^2}{a^2}\right)& = \frac{e^{ax}\big(a\cos(bx+c)+b\sin(bx+c)\big)}{a^2}\\ \Rightarrow I&=\frac{e^{ax}\big(a\cos(bx+c)+b\sin(bx+c)\big)}{a^2 + b^2}.\ _\square \end{align}\]

Note: The above example is also applicable for the form \[I = \int e^{ax}\sin(bx+c).\]

Find the indefinite integral \[\int e^{2x}\cos(5x+3)\, dx,\] using \(C\) as the constant of integration.

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By the above result, we obtain our answer as \[\frac{e^{2x}\big(2\cos(5x+3)+5\sin(5x+3)\big)}{29} + C.\ _\square\]

Case 3: If the integration is of the form \(\displaystyle \int \frac{ae^x + be^{-x}}{pe^x + qe^{-x}} dx,\) express \(\text{(NUM)} =\alpha \text{(DEN)} + \beta \frac{d}{dx} \text{(DEN)},\) where NUM=(the numerator of the integrand) and DEN=(the denominator of the integrand), and then integrate as usual.

Find the indefinite integral \[\int \frac{2e^x + 3e^{-x}}{e^x - 5e^{-x}}\, dx,\] using C as the constant of integration.

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We can write \(2e^x + 3e^{-x} = \alpha(e^x - 5e^{-x}) + \beta(e^x + 5e^{-x}).\) Comparing the coefficients of \(e^x \text{ and } e^{-x}\), we obtain \(\alpha + \beta = 2\) and \(\alpha - \beta = -\frac{3}{5},\) which implies \(\alpha = \frac7{10}, \beta=\frac{13}{10}\). So we have \[\int \frac{2e^x + 3e^{-x}}{e^x - 5e^{-x}} dx=\alpha \int dx + \beta \int \frac{e^x + 5e^{-x}}{e^x - 5e^{-x}} dx. \qquad (1)\] Let \(e^x - 5e^{-x} = t,\) then \((e^x + 5e^{-x})dx = dt,\) which gives \[\begin{align} (1) &=\frac7{10} \int dx +\frac{13}{10}\int \frac{dt}{t}\\ &=\frac{7x}{10} +\frac{13}{10}\ln |t| + C \\ &=\frac{7x}{10} +\frac{13}{10}\ln |e^x - 5e^{-x}| + C, \end{align}\] where \(C\) is the constant of integration. \(_\square\)