Integration of Exponential Functions

Exponential functions are those of the form \(f(x)=Ce^{x}\) for a constant \(C\), and the linear shifts, inverses, and quotients of such functions. Exponential functions occur frequently in physical sciences, so it can be very helpful to be able to integrate them.

Nearly all of these integrals come down to two basic formulas:

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

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:

\[\begin{align} I &= \int e^{ax}\cos(bx+c)\ dx\\ &= \cos(bx+c)\frac{e^{ax}}{a} + \frac{b}{a} \int e^{ax}\sin(bx+c)\ dx\\\\ &= \cos(bx+c)\frac{e^{ax}}{a} + \frac{b}{a}\left(\frac{e^{ax}}{a}\sin(bx+c) - \frac{b}{a} \int e^{ax}\cos(bx+c)\right)\ dx\\\\ &=\frac{e^{ax}\big(a\cos(bx+c)+b\sin(bx+c)\big)}{a^2} - \frac{b^2}{a^2}I. \end{align}\]

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 \(\displaystyle I = \int e^{ax}\sin(bx+c).\)

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

\[\begin{align} (1) &=\frac7{10} \int dx +\frac{13}{10}\int \frac{dt}{t}\\ &=\frac{7x}{10} +\frac{13}{10}\ln |t| + C \\ &=\frac{7x}{10} +\frac{13}{10}\ln \big|e^x - 5e^{-x}\big| + C, \end{align}\]

where \(C\) is the constant of integration. \(_\square\)