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Integration by parts is simply the product rule reversed. The formula is as follows:

Formula

\int u v^{'} d x = u v - \int u^{'} v d x

Note! In exercises with integration by parts, you should choose $e^{x}$ as $v^{'}$ and $\ln (x)$ as $u$ .

Example 1

\begin{array}{l} \int 3 x e^{x} d x & \overset{*}{=} 3 x e^{x} - \int 3 e^{x} d x \\ = 3 x e^{x} - 3 e^{x} + C \\ = 3 e^{x} (x - 1) + C \end{array}

\begin{aligned} u & = 3 x & v^{'} & = e^{x} \\ u^{'} & = 3 & v & = e^{x} \end{aligned}

Example 2

Find the function $F$ such that $F^{'} (x) = 4 x^{3} + \frac{1}{x}$ and $F (1) = 2$

F (x) = \int 4 x^{3} + \frac{1}{x} d x = x^{4} + \ln | x | + C

Furthermore, given that $F (1) = 2$ : $\begin{array}{l} 2 & = F (1) \\ = 1^{4} + \ln | 1 | + C \\ = 1 + 0 + C, \\ C & = 1 \end{array}$

Then, $F (x) = x^{4} + \ln | x | + 1$ .

Example 3

Compute the integral $\int \sin^{2} x d x$

\begin{array}{l} \int \sin^{2} x d x & = \int \sin x \cdot \sin x d x \\ \overset{*}{=} - \sin x \cos x \\ + \int \cos^{2} x d x \\ = - \sin x \cos x \\ + \int 1 - \sin^{2} x d x \\ = - \sin x \cos x \\ + x \int \sin^{2} x d x \end{array}

\begin{array}{l} \int \sin^{2} x d x & = \int \sin x \cdot \sin x d x \\ \overset{*}{=} - \sin x \cos x + \int \cos^{2} x d x \\ = - \sin x \cos x + \int 1 - \sin^{2} x d x \\ = - \sin x \cos x + x - \int \sin^{2} x d x \end{array}

\begin{aligned} u & = \sin x & v^{'} & = \sin x \\ u^{'} & = \cos x & v & = - \cos x \end{aligned}

This gives you an equation that you solve for $\int \sin^{2} x d x$ :

\begin{array}{l} \int \sin^{2} x d x & = - \sin x \cos x \\ + x \int \sin^{2} x d x \end{array}

\begin{matrix} \frac{2 \cdot \int \sin^{2} x d x}{2} = \frac{- \sin x \cos x + x}{2} \\ \int \sin^{2} x d x = - \frac{1}{2} \sin x \cos x + \frac{x}{2} + C \end{matrix}

\begin{array}{l} \int \sin^{2} x d x & = - \sin x \cos x + x - \int \sin^{2} x d x \\ 2 \cdot \int \sin^{2} x d x & = - \sin x \cos x + x | : 2 \\ \int \sin^{2} x d x & = - \frac{1}{2} \sin x \cos x + \frac{x}{2} + C \end{array}

Example 4

Compute $\int \cos (2 x) \sin (2 x) d x$

\begin{array}{l} \int \cos (2 x) \sin (2 x) d x \\ \overset{*}{=} - \frac{1}{2} \cos^{2} (2 x) - \int \sin (2 x) \cos (2 x) d x \end{array}

\int \cos (2 x) \sin (2 x) d x \overset{*}{=} - \frac{1}{2} \cos^{2} (2 x) - \int \sin (2 x) \cos (2 x) d x

\begin{aligned} u & = \cos (2 x) & v^{'} & = \sin (2 x) \\ u^{'} & = - 2 \sin (2 x) & v & = - \frac{1}{2} \cos (2 x) \end{aligned}

You now solve this expression as an equation with respect to $\int \cos (2 x) \sin (2 x) d x$ :

\begin{array}{l} \int \cos (2 x) \sin (2 x) d x & = - \frac{1}{2} \cos^{2} (2 x) \\ - \int \sin (2 x) \\ \cdot \cos (2 x) d x \\ 2 \int \cos (2 x) \sin (2 x) d x & = - \frac{1}{2} \cos^{2} (2 x) | \div 2 \\ \int \cos (2 x) \sin (2 x) d x & = - \frac{1}{4} \cos^{2} (2 x) + C \end{array}

\begin{array}{l} \int \cos (2 x) \sin (2 x) d x & = - \frac{1}{2} \cos^{2} (2 x) - \int \sin (2 x) \cos (2 x) d x \\ 2 \int \cos (2 x) \sin (2 x) d x & = - \frac{1}{2} \cos^{2} (2 x) | \div 2 \\ \int \cos (2 x) \sin (2 x) d x & = - \frac{1}{4} \cos^{2} (2 x) + C \end{array}

Example 5

Compute $\int e^{x} (x^{2} + 3 x - 4) d x$

\begin{array}{l} \int e^{x} (x^{2} + 3 x - 4) d x \\ \overset{*}{=} e^{x} (x^{2} + 3 x - 4) - \int e^{x} (2 x + 3) d x \\ = e^{x} (x^{2} + 3 x - 4) \\ - (e^{x} (2 x + 3) - \int 2 e^{x} d x) \\ \overset{* *}{=} e^{x} (x^{2} + 3 x - 4) \\ - e^{x} (2 x + 3) + 2 e^{x} + C \\ = e^{x} (x^{2} + x - 5) + C \end{array}

\begin{array}{l} \int e^{x} (x^{2} + 3 x - 4) d x & \overset{*}{=} e^{x} (x^{2} + 3 x - 4) - \int e^{x} (2 x + 3) d x \\ = e^{x} (x^{2} + 3 x - 4) - (e^{x} (2 x + 3) - \int 2 e^{x} d x) \\ \overset{* *}{=} e^{x} (x^{2} + 3 x - 4) - e^{x} (2 x + 3) + 2 e^{x} + C \\ = e^{x} (x^{2} + x - 5) + C \end{array}

\begin{aligned} u & = x^{2} + 3 x - 4 & v^{'} & = e^{x} \\ u^{'} & = 2 x + 3 & v & = e^{x} \end{aligned}

\begin{aligned} z & = 2 x + 3 & w^{'} & = e^{x} \\ z^{'} & = 2 & w & = e^{x} \end{aligned}

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How to Use the Formula for Integration by Parts