Integration by Substitution

Method of Integration by Substitution

Method of Integration by Substitution

1. Identify a suitable substitution, \(u=g(x)\), by rewriting the integral in one of the following two forms:

   \[\int [g(x)]^n g'(x) ~dx \qquad \text{or} \qquad \int e^{g(x)} g'(x) ~dx\]

   and compute the corresponding differential \(du= g'(x) ~dx\).

2. Rewrite the integral in terms of \(u\) and \(du\). To complete the substitution, it may be helpful to divide both sides of \(du= g'(x) ~dx\) by a constant and/or to solve for \(x\) in terms of \(u\) in the equation \(u=g(x)\).

3. Evaluate the integral in terms of \(u\).

4. Replace \(u\) with \(g(x)\).

Example 1

Compute \(\displaystyle \int 20x^3 \sqrt{5x^4+7} ~dx\).

Step 1: Identify a suitable substitution.

Based on rewriting the integral in the following form

\[\int (5x^4+7)^{1/2} ~20x^3 ~dx\]

let \(u=5x^4 + 7\) and \(du = 20x^3 ~dx\).

Step 2: Rewrite the integral in terms of \(u\) and \(du\).

\[\int (5x^4+7)^{1/2} ~20x^3 ~dx = \int u^{1/2} ~du\]

Step 3: Evaluate the integral in terms of \(u\).

\[\begin{align*} \int u^{1/2} ~du &= \frac{u^{1/2+1}}{1/2+1} + C &&\text{Apply the Power Rule}\\ &= \frac{u^{3/2}}{3/2} + C &&\text{Simplify}\\ &= \frac{2}{3}u^{3/2} + C && \text{Simplify} \end{align*}\]

Step 4: Replace \(u\) with \(5x^4 + 7\).

\[ \int 20x^3 \sqrt{5x^4+7} ~dx ~=~ \frac{2}{3}(5x^4 + 7)^{3/2} + C \]

Example 2

Compute \(\displaystyle \int 7x^2e^{4x^3+5} ~dx\).

Step 1: Identify a suitable substitution.

Based on rewriting the integral in the following form

\[7\int e^{4x^3+5} ~x^2 ~dx\]

let \(u=4x^3 + 5\) and \(du = 12x^2 ~dx\), or equivalently \(\dfrac{1}{12}du = x^2 ~dx\).

Step 2: Rewrite the integral in terms of \(u\) and \(du\).

\[\begin{align*} 7\int e^{4x^3+5} ~x^2 ~dx &= 7\int e^{u} \frac{1}{12} ~du \\ &= \frac{7}{12}\int e^{u} ~du \end{align*}\]

Step 3: Evaluate the integral in terms of \(u\).

\[\begin{align*} \frac{7}{12}\int e^{u} ~du &= \frac{7}{12}e^u + C \end{align*}\]

Step 4: Replace \(u\) with \(4x^3 + 5\).

\[ \int 7x^2e^{4x^3+5} ~dx ~=~ \frac{7}{12}e^{4x^3 + 5} + C \]

Example 3

Compute \(\displaystyle \int \frac{(\ln x)^2}{x}~dx\).

Step 1: Identify a suitable substitution.

Based on rewriting the integral in the following form

\[\int (\ln x)^2\frac{1}{x} ~dx\]

let \(u=\ln x\) and \(du = \dfrac{1}{x} ~dx\).

Step 2: Rewrite the integral in terms of \(u\) and \(du\).

\[\begin{align*} \int (\ln x)^2\frac{1}{x} ~dx &= \int u^2 ~du \end{align*}\]

Step 3: Evaluate the integral in terms of \(u\).

\[\begin{align*} \int u^2 ~du &= \frac{1}{3}u^3 + C \end{align*}\]

Step 4: Replace \(u\) with \(\ln x\).

\[ \int \frac{(\ln x)^2}{x}~dx ~=~ \frac{1}{3}(\ln x)^3 + C \]

Example 4

Compute \(\displaystyle \int \frac{x}{x+4} ~dx\).

Step 1: Identify a suitable substitution.

Based on rewriting the integral in the following form

\[\int (x+4)^{-1} x ~dx\]

let \(u=x+4\) and \(du = dx\).

Step 2: Rewrite the integral in terms of \(u\) and \(du\).

In this case, it is necessary to solve for \(x\) in terms of \(u\) in the equation \(u = x+4\) (i.e., \(x = u-4\)) to complete the substitution.

\[\begin{align*} \int (x+4)^{-1} x ~dx &= \int u^{-1} (u-4) ~du \\ &= \int \frac{u-4}{u} ~du \end{align*}\]

Step 3: Evaluate the integral in terms of \(u\).

\[\begin{align*} \int \frac{u-4}{u} ~du &= \int \frac{u}{u}- \frac{4}{u} ~du && \hbox{Rewrite integrand as a sum}\\ &= \int 1- \frac{4}{u} ~du && \hbox{Simplify}\\ &= \int 1 ~du - 4\int \frac{1}{u} ~du && \hbox{Sum & constant multiple rules}\\ &= u - 4\ln|u| + C && \hbox{Integrate each term} \end{align*}\]

Step 4: Replace \(u\) with \(x+4\).

\[\begin{align*} \int \frac{x}{x+4} ~dx &= x+4 - 4\ln|x+4| +C \\ &= x - 4\ln|x+4| +C + 4 \\ &= x - 4\ln|x+4| +C \end{align*}\]

Observation

The last line follows from the observation that if \(C\) is an arbitrary constant, then so is \(C+4\). In other words, the \(+4\) can be absorbed into the arbitrary constant \(C\), and is not needed. We can verify this final answer by computing its derivative.

\[\begin{align*} \frac{d}{dx}(x - 4\ln|x+4|) &= 1 - 4\frac{1}{x+4} \\ &= \frac{x+4}{x+4} - \frac{4}{x+4} && \hbox{Get a common denominator}\\ &= \frac{x+4 - 4}{x+4}&& \hbox{Combine numerators} \\ &= \frac{x}{x+4} && \hbox{Simplify} \end{align*}\]

Therefore, \(x - 4\ln|x+4|\) is an antiderivative of \(\dfrac{x}{x+4}\).