Basic Rules of Integration

Basic Rules of Integration#

The Rules#

Integral of a Costant

For any real number \(k\),

\[\int k ~dx = kx + C\]

The Power Rule

For any real number \(n\neq -1\),

\[\int x^n ~dx = \frac{x^{n+1}}{n+1} + C\]

The Constant Multiple Rule

For any real number \(k\),

\[\int kf(x) ~dx = k\int f(x) ~dx\]

The Sum Rule

\[\int f(x)\pm g(x) ~dx = \int f(x) ~dx \pm \int g(x) ~dx\]

Integral of \(e^x\)

For any real number \(a\neq 0\),

\[\int e^x ~dx = e^x + C ~~~~~ \text{and} ~~~~~ \int e^{ax} ~dx = \frac{e^{ax}}{a} + C\]

Integral of \(1/x\)

\[\int \frac{1}{x} ~dx = \ln|x| + C\]

The Substitution Rule

If \(u=g(x)\) and \(du = g'(x) ~dx\) then

\[\int f(g(x))g'(x) ~dx = \int f(u) ~du\]

Example 1#

Integral of a constant

Compute \(\displaystyle \int 13 ~dz\).

Step 1:   Notice the differential \(dz\).

This indicates that we are looking for a function of \(z\), \(f(z)\), such that \(f'(z) = 13\).

Step 2:   Recall the formula for the integral of a constant.
\[\int k ~dx = kx + C\]
Step 3:   Apply the rule with \(k = 13\).

Therefore,

\[\int 13 ~dz = 13z + C\]

In other words, every function of \(z\) with derivative equal to \(13\) can be written in the form \(13z + C\) for some constant \(C\).

Check Our Work.

We can verify our answer by showing \(13\) is the derivative of \(13z\).

\[\frac{d}{dz} 13z = 13\]

Example 2#

Integral of a power function

Compute \(\displaystyle \int x^7 ~dx\).

Step 1:   Recall the power rule.

For any real number \(n\neq -1\),

\[\int x^n ~dx = \frac{x^{n+1}}{n+1} + C\]
Step 2:   Apply the power rule with \(n = 7\).
\[\begin{align*} \int x^7 dx &= \frac{x^{7+1}}{7+1} + C \\ &= \frac{x^8}{8} + C \end{align*}\]
Check Our Work.

We can verify our answer by showing \(x^7\) is the derivative of \(\dfrac{x^8}{8}\).

\[\begin{align*} \frac{d}{dx} \frac{x^8}{8} &= \frac{d}{dx} \frac{1}{8}x^8 \\ &= \frac{1}{8} \frac{d}{dx} x^8 \\ &= \frac{1}{8} 8x^7 \\ &= x^7 \end{align*}\]

Example 3#

Integral of a power function with rational exponent

Compute \(\displaystyle \int \frac{1}{\sqrt{y}} ~dy\).

Step 1:   Rewrite the integrand in the appropriate form to apply the power rule.
\[\int \frac{1}{\sqrt{y}} ~dy = \int y^{-1/2} ~dy\]
Step 2:   Apply the power rule with \(n = -1/2\).
\[\begin{align*} \int y^{-1/2} ~dy &= \frac{y^{-1/2 + 1}}{-1/2 + 1} + C \\ &= \frac{y^{1/2}}{1/2} + C \\ &= 2\sqrt{y} + C \end{align*}\]
Check Our Work.

We can verify our answer by showing \(\dfrac{1}{\sqrt{y}}\) is the derivative of \(2\sqrt{y}\).

\[\begin{align*} \frac{d}{dy} 2\sqrt{y} &= 2 \frac{d}{dy} y^{1/2} \\ &= 2 \frac{1}{2} y^{-1/2} \\ &= \dfrac{1}{\sqrt{y}} \end{align*}\]

Example 4#

Integral of a constant multiple of a power function

Compute \(\displaystyle \int 4x^{7/3} ~dx\).

Step 1:   Recall the constant multiple rule.

For any real number \(k\),

\[\int kf(x) ~dx = k\int f(x) ~dx\]
Step 2:   Apply the constant multiple rule with \(k=4\).
\[\begin{align*} \int 4x^{7/3} ~dx &= 4\int x^{7/3} ~dx \\ &= 4 \left(\frac{x^{7/3+1}}{7/3+1}\right) + C && \text{power rule with $n=7/3$}\\ &= 4 \left( \frac{x^{10/3}}{10/3}\right) + C && \hbox{simplify} \\ &= 4 \cdot \frac{3}{10} x^{10/3} + C \\ &= \frac{6}{5} x^{10/3} + C \end{align*}\]
Check Our Work.

We can verify our answer by showing \(4x^{7/3}\) is the derivative of \(\dfrac{6}{5} x^{10/3}\).

\[\begin{align*} \frac{d}{dx} \frac{6}{5} x^{10/3} &= \frac{6}{5} \frac{d}{dx} x^{10/3} \\ &= \frac{6}{5} \cdot \frac{10}{3} x^{7/3} \\ &= 4x^{7/3} \end{align*}\]

Example 5#

Integral of a sum of power functions

Compute \(\displaystyle \int 5t^3-\frac{10}{t^{6}}+4\sqrt{t} ~dt\).

Step 1:   Recall the sum and difference rule.
\[\int f(x)\pm g(x) ~dx = \int f(x) ~dx \pm \int g(x) ~dx\]
Step 2:   Apply the sum, difference and constant multiple rules.
\[\begin{align*} \int 5t^3-\frac{10}{t^{6}}+4\sqrt{t} ~dt &= \int 5t^3 ~dt - \int \frac{10}{t^{6}} ~dt + \int 4\sqrt{t} ~dt && \text{sum rule}\\ &= 5\int t^3 ~dt - 10\int \frac{1}{t^{6}} ~dt + 4\int \sqrt{t} ~dt && \text{constant multiple rule}\\ &= 5\int t^3 ~dt - 10\int t^{-6} ~dt + 4\int t^{1/2} ~dt \end{align*}\]
Step 3:   Integrate each term and then simplify.
\[\begin{align*} 5\int t^3 ~dt - 10\int t^{-6} ~dt + 4\int t^{1/2} ~dt &= 5\cdot \frac{t^4}{4} - 10\cdot \frac{t^{-5}}{-5} + 4\frac{t^{3/2}}{3/2} + C && \hbox{power rule}\\ &= \frac{5}{4}t^4 + 2t^{-5} + \frac{8}{3}t^{3/2} + C && \hbox{simplify} \end{align*}\]
Check Our Work.

We can verify our answer by showing \(5t^3-\dfrac{10}{t^{6}}+4\sqrt{t}\) is the derivative of \(\dfrac{5}{4}t^4 + 2t^{-5} + \dfrac{8}{3}t^{3/2}\).

\[\begin{align*} \frac{d}{dt} \left(\frac{5}{4}t^4 + 2t^{-5} + \frac{8}{3}t^{3/2} \right) &= \frac{5}{4}4t^3 + 2(-5)t^{-6} + \frac{8}{3}\cdot \frac{3}{2}t^{1/2} \\ &= 5t^3-\frac{10}{t^{6}}+4\sqrt{t} \end{align*}\]

Example 6#

Integral of a polynomial divided by a power function

Compute \(\displaystyle \int \frac{4x^9 - 15x^4 + 7x^3}{x^4} ~dx\).

Step 1:   Rewrite the integrand as a sum.
\[\begin{align*} \frac{4x^9-15x^4 + 7x^3}{x^4} &= \frac{4x^9}{x^4} - \frac{15x^4}{x^4} + \frac{7x^3}{x^4}\\ &= 4x^5 - 15 + \frac{7}{x} \end{align*}\]
Step 2:   Apply the sum and constant multiple rules.
\[\begin{align*} \int \frac{4x^9 - 15x^4 + 7x^3}{x^4} ~dx &= \int 4x^5 - 15 + \frac{7}{x} ~dx \\ \\ &= \int 4x^5 ~dx - \int 15 ~dx + \int \frac{7}{x}~dx && \hbox{sum rule}\\ \\ &= 4\int x^5 ~dx - \int 15 ~dx + 7\int \frac{1}{x}~dx && \hbox{constant multiple rule} \end{align*}\]
Step 3:   Integrate each term and then simplify.
\[\begin{align*} 4\int x^5 ~dx - \int 15 ~dx + 7\int \frac{1}{x}~dx &= \frac{4x^6}{6} - 15x + 7\ln|x| + C && \text{integrate each term}\\ \\ &= \frac{2x^6}{3} - 15x + 7\ln|x| + C && \text{simplify}\\ \end{align*}\]

Example 7#

Integral of an exponential function

Compute \(\displaystyle \int e^{2x/5} ~dx\).

Step 1:   Recall the formula for the integral of \(e^{ax}\) for \(a\neq 0\).
\[\int e^{ax} ~dx = \frac{e^{ax}}{a} + C\]
Step 2:   Apply the formula for the integral of \(e^{ax}\) with \(a=2/5\).
\[\begin{align*} \int e^{2x/5} ~dx &= \frac{e^{2x/5}}{2/5} + C\\ \\ &= \frac{5}{2}e^{2x/5} + C \end{align*}\]

Example 8#

Integral of a sum of functions

Compute \(\displaystyle \int 3e^{2x}+\frac{8}{x}+\frac{4}{x^3} ~dx\).

Step 1:   Apply the sum and constant multiple rules.
\[\begin{align*} \int 3e^{2x}+\frac{8}{x}+\frac{4}{x^3} ~dx &= \int 3e^{2x} ~dx + \int \frac{8}{x} ~dx + \int 4x^{-3}~dx && \hbox{sum rule}\\ \\ &= 3\int e^{2x} ~dx + 8\int \frac{1}{x} ~dx + 4\int x^{-3}~dx && \hbox{constant multiple rule} \end{align*}\]
Step 2:   Integrate each term and then simplify.
\[\begin{align*} &= 3\frac{e^{2x}}{2} + 8\int \frac{1}{x} ~dx + 4\int x^{-3}dx && \text{integral of $e^{ax}$}\\ \\ &= 3\frac{e^{2x}}{2} + 8\ln|x| + 4\int x^{-3}dx && \text{integral of $1/x$}\\ \\ &= 3\frac{e^{2x}}{2} + 8\ln|x| + \frac{4x^{-2}}{-2}+C && \text{power rule}\\ \\ &=\frac{3}{2}e^{2x} + 8\ln|x| - \frac{2}{x^{2}}+C && \text{simplify} \end{align*}\]

Example 9#

Integral of a product of functions

Compute \(\displaystyle \int (e^{3x} + 1)(e^{-3x} - 1) ~dx\).

Step 1:   Rewrite the integrand as a sum.
\[\begin{align*} (e^{3x} + 1)(e^{-3x} - 1) &= e^{3x}e^{-3x} + e^{-3x} - e^{3x} - 1 && \hbox{FOIL}\\ &= e^{3x-3x} + e^{-3x} - e^{3x} - 1 && \hbox{since $e^a e^b = e^{a+b}$}\\ &= e^{0} + e^{-3x} - e^{3x} - 1\\ &= 1 + e^{-3x} - e^{3x} - 1 && \hbox{since $e^0 = 1$}\\ &= e^{-3x} - e^{3x} && \hbox{simplify} \end{align*}\]
Step 2:   Apply the sum rule and then integrate each term.
\[\begin{align*} \int (e^{3x} + 1)(e^{-3x} - 1) ~dx &= \int e^{-3x} - e^{3x} ~dx \\ &= \int e^{-3x} ~dx - \int e^{3x} ~dx \\ &= \frac{1}{-3} e^{-3x} - \frac{1}{3} e^{3x} + C && \text{integral of $e^{ax}$}\\ &= -\frac{1}{3}(e^{-3x} + e^{3x}) + C && \text{simplify} \end{align*}\]