Integration by Parts of Indefinite Integrals Examples 1

# Integration by Parts of Indefinite Integrals Examples 1

Recall that if we have a function containing two parts to which we can split up and assign one of them "$u$" and the other "$dv$, then it thus follows that:

(1)
\begin{align} \int u \: dv = uv - \int v \: du \end{align}

We will now apply this formula in the following examples.

## Example 1

Evaluate the following integral: $\int x^2 \ln x \: dx$.

We let $u = \ln x$ since when differentiated, it becomes simpler and more beneficial than $x^2$, and $dv = x^2 \: dx$. It thus follows that $du = 1/x \: dx$ and $v = x^3/3$. Substituting this into the integration by parts formula we obtain that:

(2)
\begin{align} \int x^2 \ln x \: dx = \frac{x^3 \ln x}{3} - \int \frac{x^3}{3} \cdot \frac{1}{x} \: dx \\ \int x^2 \ln x \: dx = \frac{x^3 \ln x}{3} - \frac{1}{3} \int x^2 \: dx \\ \int x^2 \ln x \: dx = \frac{x^3 \ln x}{3} - \frac{1}{9} x^3 + C \\ \end{align}

## Example 2

Evaluate the following integral: $\int x \cos {5x} \: dx$.

We select $u = x$ since when differentiated, it becomes simpler, and $du = \cos {5x} \: dx$. It thus follows that $du = dx$ and $v = \frac{1}{5} \sin {5x}$. Substituting this into the integration by parts formula and we obtain that:

(3)
\begin{align} \int x \cos {5x} \: dx = \frac{x \sin x}{5} - \int \frac{1}{5} \sin{5x} \: dx \\ \int x \cos {5x} \: dx = \frac{x \sin x}{5} - \frac{1}{5} \int \sin {5x} \: dx \\ \int x \cos {5x} \: dx = \frac{x \sin x}{5} + \frac{\cos {5x}}{25} + C \end{align}

## Example 3

Evaluate the following integral: $\int x e^{x/2} \: dx$.

First let $u = x$ since when differentiated it becomes simpler, and $dv = e^{x/2} \: dx$. It thus follows that $du = dx$ and $v = 2e^{x/2}$. Substituting this into the integration by parts formula and we obtain that:

(4)
\begin{align} \int x e^{x/2} \: dx = 2xe^{x/2} - \int 2e^{x/2} \: dx \\ \int x e^{x/2} \: dx = 2xe^{x/2} - 2 \int e^{x/2} \: dx \\ \int x e^{x/2} \: dx = 2xe^{x/2} - 4 e^{x/2} + C \\ \int x e^{x/2} \: dx = 2e^{x/2}(x - 2) + C \end{align}

## Example 4

Evaluate the following integral: $\int x^2 e^x \: dx$.

We note that $e^x$ remains unchanged when differentiated, so we let $u = x^2$ and $dv = e^x \: dx$. Therefore, $du = 2x \: dx$ and $v = e^x$. Applying the integration by parts formula we obtain that:

(5)
\begin{align} \int x^2 e^x \: dx = x^2 e^x - \int 2x e^x \: dx \\ \int x^2 e^x \: dx = x^2 e^x - 2\int x e^x \: dx \end{align}

We must now apply integration by parts once again for the rightmost integral. Let's choose $u = x$ and $dv = e^x \: dx$. Therefore, $du = dx$ and $v = e^x$. Substituting this back, we get:

(6)
\begin{align} \int x^2 e^x \: dx = x^2 e^x - 2\left ( xe^x - \int e^x \: dx \right ) \\ \int x^2 e^x \: dx = x^2 e^x - 2\left ( xe^x - e^x \right ) + C \end{align}