The easiest expressions to integrate are polynomials. I'll reference constants included in the integration with \(k\) and constants of integration as \(c\).

As a quick refresher, whenever we evaluate an

indefinite integral, meaning the general form and withoutlimits of integration, we always include a constant in the final answer since the derivative of any constant is \(0\).

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Notice that this is a special case of the following *power rule* when \(n = 0\).

When integrating any \(x^n\) where \(n \neq -1\),

\[ \int kx^n\ dx = \frac{k}{n+1} x^{n+1} + c \,.\]Integrating the reciprocal of \(x\) yields the logarithmic function,

\[ \int \frac{1}{kx}\ dx = \frac{1}{k}\int (x)^{-1}\ dx = \frac{1}{k}\log |x| + c\,.\]The exponential function is simultaneously the most boring and most interesting function of all.

\[ \int e^{\pm kx}\ dx = \pm \frac{1}{k} e^{\pm kx} + c\,.\]The Taylor Series expansion of the exponential function, which is defined for all reals, \(\mathbb{R}\), is,

\[ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots\,,\]or more compactly written as,

\[ e^x = \sum_{n=0}^\infty x^n\,. \]Table of common trig integrations coming soon...

Given a function such as, \(f(x) = \sqrt{k_1 + k_2 x }\), we can perform a simple substitution such as, \(u = k_1 + k_2 x\). Since we'll also need to transform the \(dx\) variable as well, we differentiate \(u\) to access \(dx\),

\[ \begin{aligned} u &= k_1 + k_2 x \\ du &= k_2\ dx \implies dx = \frac{1}{k_2}\ du \end{aligned}\]Therefore, plugging \((7)\) into the integration,

\[\int \sqrt{k_1 + k_2 x }\ dx \rightarrow \frac{1}{k_2}\int \sqrt{u}\ du\,, \]and since roots are just fractional exponents, we apply the power rule and then substitute the original function back in for \(u\),

\[ \begin{aligned}\frac{1}{k_2}\int u^{1/2}\ du &= \frac{1}{k_2}\frac{1}{\frac{3}{2}} u^{3/2} + c \\ &= \frac{2}{3k_2} u^{3/2} + c \\ &= \frac{2}{3k_2}\sqrt{(k_1 + k_2 x)^3} + c \end{aligned}\]Don't forget to account for all of the variables! In this case, when we differentiated, all of the \(x\) variables went away and we are perfectly fine to move constants around in the integrand. However, suppose that inside of the square root was an \(x^n\) term where \(n \geq 2\), then when differentiating we would still have an \(x\) term left over. Since \(x\) is the variable we are integrating with respect to, that wouldn't work. Suppose that the original function was \(f(x) = x\sqrt{k_1 + k_2 x^2}\). In this case, we could set \(u = k_1 + k_2 x^2\), differentiate to get \(du = 2 k_2 x\ dx\) and this would work fine because we can equate the left over \(x\) with the \(x\) in front of the square root and all is right in the world.

Suppose \(u\) and \(v\) are functions are \(x\).

\[ \frac{d}{dx}(u\cdot v) = u\frac{d}{dx}v + v\frac{d}{dx}u \] \[\begin{aligned} \int \frac{d}{dx}(u\cdot v)\ dx &= \int u\frac{d}{dx}v\ dx + \int v\frac{d}{dx}u\ dx \\ \int u\frac{d}{dx}v\ dx + \int v\frac{d}{dx}u\ dx &= \int \frac{d}{dx}(u\cdot v)\ dx \\ \int u\ dv &= u\cdot v - \int v\ du \\ \end{aligned}\]As an example, suppose we are evaluating, \(\int e^x \cos(x)\ dx\).

Follow the mnemonic, LIATE (Logarithms, Inverse Trig, Algebraic, Trig, and Exponential) for selecting \(u\) and \(dv\).

\[\begin{aligned} u &= \cos(x) \\ du &= -\sin(x)\ dx \\ dv &= e^x\ dx\\ v &= e^x \\ \end{aligned}\] \[\int e^x \cos(x)\ dx = e^x \cos(x) + \int e^x \sin(x)\ dx \]We need to play the same game with the \(v\ du\) term.

\[\begin{aligned} u &= \sin(x) \\ du &= \cos(x)\ dx \\ dv &= e^x\ dx\\ v &= e^x \\ \end{aligned}\] \[\int e^x \sin(x)\ dx = e^x \sin(x) - \int e^x \cos(x)\ dx \] \[\begin{aligned} \int e^x \cos(x)\ dx &= e^x \cos(x) + \Big[ e^x \sin(x) - \int e^x \cos(x)\ dx \Big]\\ &= e^x \cos(x) + e^x \sin(x) - \int e^x \cos(x)\ dx \end{aligned}\]Let \(I = \int e^x \cos(x)\ dx\), then

\[\begin{aligned} I &= e^x \cos(x) + e^x \sin(x) - I\\ 2I &= e^x \cos(x) + e^x \sin(x) \\ \\ \int e^x \cos(x)\ dx &= \frac{e^x}{2}\Big(\cos(x) + \sin(x)\Big) + c \end{aligned}\]
© Daniel Marvin. Last modified: July 19, 2021.