Calculus Examples

$\int \frac{5 x + 1}{(2 x + 1) (x - 1)} d x$

Step 1

Write the fraction using partial fraction decomposition.

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Step 1.1

Decompose the fraction and multiply through by the common denominator.

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Step 1.1.1

For each factor in the denominator, create a new fraction using the factor as the denominator, and an unknown value as the numerator. Since the factor in the denominator is linear, put a single variable in its place $A$ .

$\frac{A}{2 x + 1}$

Step 1.1.2

$\frac{A}{2 x + 1} + \frac{B}{x - 1}$

Step 1.1.3

Multiply each fraction in the equation by the denominator of the original expression. In this case, the denominator is $(2 x + 1) (x - 1)$ .

$\frac{(5 x + 1) (2 x + 1) (x - 1)}{(2 x + 1) (x - 1)} = \frac{(A) (2 x + 1) (x - 1)}{2 x + 1} + \frac{(B) (2 x + 1) (x - 1)}{x - 1}$

Step 1.1.4

Cancel the common factor of $2 x + 1$ .

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Step 1.1.4.1

Cancel the common factor.

$\frac{(5 x + 1) (2 x + 1) (x - 1)}{(2 x + 1) (x - 1)} = \frac{(A) (2 x + 1) (x - 1)}{2 x + 1} + \frac{(B) (2 x + 1) (x - 1)}{x - 1}$

Step 1.1.4.2

Rewrite the expression.

$\frac{(5 x + 1) (x - 1)}{x - 1} = \frac{(A) (2 x + 1) (x - 1)}{2 x + 1} + \frac{(B) (2 x + 1) (x - 1)}{x - 1}$

Step 1.1.5

Cancel the common factor of $x - 1$ .

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Step 1.1.5.1

Cancel the common factor.

$\frac{(5 x + 1) (x - 1)}{x - 1} = \frac{(A) (2 x + 1) (x - 1)}{2 x + 1} + \frac{(B) (2 x + 1) (x - 1)}{x - 1}$

Step 1.1.5.2

Divide $5 x + 1$ by $1$ .

$5 x + 1 = \frac{(A) (2 x + 1) (x - 1)}{2 x + 1} + \frac{(B) (2 x + 1) (x - 1)}{x - 1}$

Step 1.1.6

Simplify each term.

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Step 1.1.6.1

Cancel the common factor of $2 x + 1$ .

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Step 1.1.6.1.1

Cancel the common factor.

$5 x + 1 = \frac{A (2 x + 1) (x - 1)}{2 x + 1} + \frac{(B) (2 x + 1) (x - 1)}{x - 1}$

Step 1.1.6.1.2

Divide $(A) (x - 1)$ by $1$ .

$5 x + 1 = (A) (x - 1) + \frac{(B) (2 x + 1) (x - 1)}{x - 1}$

Step 1.1.6.2

Apply the distributive property.

$5 x + 1 = A x + A \cdot - 1 + \frac{(B) (2 x + 1) (x - 1)}{x - 1}$

Step 1.1.6.3

Move $- 1$ to the left of $A$ .

$5 x + 1 = A x - 1 \cdot A + \frac{(B) (2 x + 1) (x - 1)}{x - 1}$

Step 1.1.6.4

Rewrite $- 1 A$ as $- A$ .

$5 x + 1 = A x - A + \frac{(B) (2 x + 1) (x - 1)}{x - 1}$

Step 1.1.6.5

Cancel the common factor of $x - 1$ .

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Step 1.1.6.5.1

Cancel the common factor.

$5 x + 1 = A x - A + \frac{(B) (2 x + 1) (x - 1)}{x - 1}$

Step 1.1.6.5.2

Divide $(B) (2 x + 1)$ by $1$ .

$5 x + 1 = A x - A + (B) (2 x + 1)$

Step 1.1.6.6

Apply the distributive property.

$5 x + 1 = A x - A + B (2 x) + B \cdot 1$

Step 1.1.6.7

Rewrite using the commutative property of multiplication.

$5 x + 1 = A x - A + 2 B x + B \cdot 1$

Step 1.1.6.8

Multiply $B$ by $1$ .

$5 x + 1 = A x - A + 2 B x + B$

Step 1.1.7

Simplify the expression.

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Step 1.1.7.1

Move $B$ .

$5 x + 1 = A x - A + 2 x B + B$

Step 1.1.7.2

Move $- A$ .

$5 x + 1 = A x + 2 x B - A + B$

Step 1.2

Create equations for the partial fraction variables and use them to set up a system of equations.

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Step 1.2.1

Create an equation for the partial fraction variables by equating the coefficients of $x$ from each side of the equation. For the equation to be equal, the equivalent coefficients on each side of the equation must be equal.

$5 = A + 2 B$

Step 1.2.2

Create an equation for the partial fraction variables by equating the coefficients of the terms not containing $x$ . For the equation to be equal, the equivalent coefficients on each side of the equation must be equal.

$1 = - 1 A + B$

Step 1.2.3

Set up the system of equations to find the coefficients of the partial fractions.

$5 = A + 2 B$

$1 = - 1 A + B$

$5 = A + 2 B$

$1 = - 1 A + B$

Step 1.3

Solve the system of equations.

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Step 1.3.1

Solve for $A$ in $5 = A + 2 B$ .

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Step 1.3.1.1

Rewrite the equation as $A + 2 B = 5$ .

$A + 2 B = 5$

$1 = - 1 A + B$

Step 1.3.1.2

Subtract $2 B$ from both sides of the equation.

$A = 5 - 2 B$

$1 = - 1 A + B$

$A = 5 - 2 B$

$1 = - 1 A + B$

Step 1.3.2

Replace all occurrences of $A$ with $5 - 2 B$ in each equation.

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Step 1.3.2.1

Replace all occurrences of $A$ in $1 = - 1 A + B$ with $5 - 2 B$ .

$1 = - 1 (5 - 2 B) + B$

$A = 5 - 2 B$

Step 1.3.2.2

Simplify the right side.

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Step 1.3.2.2.1

Simplify $- 1 (5 - 2 B) + B$ .

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Step 1.3.2.2.1.1

Simplify each term.

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Step 1.3.2.2.1.1.1

Apply the distributive property.

$1 = - 1 \cdot 5 - 1 (- 2 B) + B$

$A = 5 - 2 B$

Step 1.3.2.2.1.1.2

Multiply $- 1$ by $5$ .

$1 = - 5 - 1 (- 2 B) + B$

$A = 5 - 2 B$

Step 1.3.2.2.1.1.3

Multiply $- 2$ by $- 1$ .

$1 = - 5 + 2 B + B$

$A = 5 - 2 B$

$1 = - 5 + 2 B + B$

$A = 5 - 2 B$

Step 1.3.2.2.1.2

Add $2 B$ and $B$ .

$1 = - 5 + 3 B$

$A = 5 - 2 B$

$1 = - 5 + 3 B$

$A = 5 - 2 B$

$1 = - 5 + 3 B$

$A = 5 - 2 B$

$1 = - 5 + 3 B$

$A = 5 - 2 B$

Step 1.3.3

Solve for $B$ in $1 = - 5 + 3 B$ .

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Step 1.3.3.1

Rewrite the equation as $- 5 + 3 B = 1$ .

$- 5 + 3 B = 1$

$A = 5 - 2 B$

Step 1.3.3.2

Move all terms not containing $B$ to the right side of the equation.

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Step 1.3.3.2.1

Add $5$ to both sides of the equation.

$3 B = 1 + 5$

$A = 5 - 2 B$

Step 1.3.3.2.2

Add $1$ and $5$ .

$3 B = 6$

$A = 5 - 2 B$

$3 B = 6$

$A = 5 - 2 B$

Step 1.3.3.3

Divide each term in $3 B = 6$ by $3$ and simplify.

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Step 1.3.3.3.1

Divide each term in $3 B = 6$ by $3$ .

$\frac{3 B}{3} = \frac{6}{3}$

$A = 5 - 2 B$

Step 1.3.3.3.2

Simplify the left side.

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Step 1.3.3.3.2.1

Cancel the common factor of $3$ .

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Step 1.3.3.3.2.1.1

Cancel the common factor.

$\frac{3 B}{3} = \frac{6}{3}$

$A = 5 - 2 B$

Step 1.3.3.3.2.1.2

Divide $B$ by $1$ .

$B = \frac{6}{3}$

$A = 5 - 2 B$

$B = \frac{6}{3}$

$A = 5 - 2 B$

$B = \frac{6}{3}$

$A = 5 - 2 B$

Step 1.3.3.3.3

Simplify the right side.

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Step 1.3.3.3.3.1

Divide $6$ by $3$ .

$B = 2$

$A = 5 - 2 B$

$B = 2$

$A = 5 - 2 B$

$B = 2$

$A = 5 - 2 B$

$B = 2$

$A = 5 - 2 B$

Step 1.3.4

Replace all occurrences of $B$ with $2$ in each equation.

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Step 1.3.4.1

Replace all occurrences of $B$ in $A = 5 - 2 B$ with $2$ .

$A = 5 - 2 \cdot 2$

$B = 2$

Step 1.3.4.2

Simplify the right side.

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Step 1.3.4.2.1

Simplify $5 - 2 \cdot 2$ .

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Step 1.3.4.2.1.1

Multiply $- 2$ by $2$ .

$A = 5 - 4$

$B = 2$

Step 1.3.4.2.1.2

Subtract $4$ from $5$ .

$A = 1$

$B = 2$

$A = 1$

$B = 2$

$A = 1$

$B = 2$

$A = 1$

$B = 2$

Step 1.3.5

List all of the solutions.

$A = 1, B = 2$

Step 1.4

Replace each of the partial fraction coefficients in $\frac{A}{2 x + 1} + \frac{B}{x - 1}$ with the values found for $A$ and $B$ .

$\frac{1}{2 x + 1} + \frac{2}{x - 1}$

Step 1.5

Remove the zero from the expression.

$\int \frac{1}{2 x + 1} + \frac{2}{x - 1} d x$

Step 2

Split the single integral into multiple integrals.

$\int \frac{1}{2 x + 1} d x + \int \frac{2}{x - 1} d x$

Step 3

Let $u_{1} = 2 x + 1$ . Then $d u_{1} = 2 d x$ , so $\frac{1}{2} d u_{1} = d x$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

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Step 3.1

Let $u_{1} = 2 x + 1$ . Find $\frac{d u_{1}}{d x}$ .

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Step 3.1.1

Differentiate $2 x + 1$ .

$\frac{d}{d x} [2 x + 1]$

Step 3.1.2

By the Sum Rule, the derivative of $2 x + 1$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [1]$ .

$\frac{d}{d x} [2 x] + \frac{d}{d x} [1]$

Step 3.1.3

Evaluate $\frac{d}{d x} [2 x]$ .

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Step 3.1.3.1

Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

$2 \frac{d}{d x} [x] + \frac{d}{d x} [1]$

Step 3.1.3.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$2 \cdot 1 + \frac{d}{d x} [1]$

Step 3.1.3.3

Multiply $2$ by $1$ .

$2 + \frac{d}{d x} [1]$

Step 3.1.4

Differentiate using the Constant Rule.

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Step 3.1.4.1

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$2 + 0$

Step 3.1.4.2

Add $2$ and $0$ .

$2$

Step 3.2

Rewrite the problem using $u_{1}$ and $d u_{1}$ .

$\int \frac{1}{u_{1}} \cdot \frac{1}{2} d u_{1} + \int \frac{2}{x - 1} d x$

Step 4

Simplify.

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Step 4.1

Multiply $\frac{1}{u_{1}}$ by $\frac{1}{2}$ .

$\int \frac{1}{u_{1} \cdot 2} d u_{1} + \int \frac{2}{x - 1} d x$

Step 4.2

Move $2$ to the left of $u_{1}$ .

$\int \frac{1}{2 u_{1}} d u_{1} + \int \frac{2}{x - 1} d x$

Step 5

Since $\frac{1}{2}$ is constant with respect to $u_{1}$ , move $\frac{1}{2}$ out of the integral.

$\frac{1}{2} \int \frac{1}{u_{1}} d u_{1} + \int \frac{2}{x - 1} d x$

Step 6

The integral of $\frac{1}{u_{1}}$ with respect to $u_{1}$ is $\ln (| u_{1} |)$ .

$\frac{1}{2} (\ln (| u_{1} |) + C) + \int \frac{2}{x - 1} d x$

Step 7

Since $2$ is constant with respect to $x$ , move $2$ out of the integral.

$\frac{1}{2} (\ln (| u_{1} |) + C) + 2 \int \frac{1}{x - 1} d x$

Step 8

Let $u_{2} = x - 1$ . Then $d u_{2} = d x$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 8.1

Let $u_{2} = x - 1$ . Find $\frac{d u_{2}}{d x}$ .

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Step 8.1.1

Differentiate $x - 1$ .

$\frac{d}{d x} [x - 1]$

Step 8.1.2

By the Sum Rule, the derivative of $x - 1$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 1]$ .

$\frac{d}{d x} [x] + \frac{d}{d x} [- 1]$

Step 8.1.3

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$1 + \frac{d}{d x} [- 1]$

Step 8.1.4

Since $- 1$ is constant with respect to $x$ , the derivative of $- 1$ with respect to $x$ is $0$ .

$1 + 0$

Step 8.1.5

Add $1$ and $0$ .

$1$

Step 8.2

Rewrite the problem using $u_{2}$ and $d u_{2}$ .

$\frac{1}{2} (\ln (| u_{1} |) + C) + 2 \int \frac{1}{u_{2}} d u_{2}$

Step 9

The integral of $\frac{1}{u_{2}}$ with respect to $u_{2}$ is $\ln (| u_{2} |)$ .

$\frac{1}{2} (\ln (| u_{1} |) + C) + 2 (\ln (| u_{2} |) + C)$

Step 10

Simplify.

$\frac{1}{2} \ln (| u_{1} |) + 2 \ln (| u_{2} |) + C$

Step 11

Substitute back in for each integration substitution variable.

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Step 11.1

Replace all occurrences of $u_{1}$ with $2 x + 1$ .

$\frac{1}{2} \ln (| 2 x + 1 |) + 2 \ln (| u_{2} |) + C$

Step 11.2

Replace all occurrences of $u_{2}$ with $x - 1$ .

$\frac{1}{2} \ln (| 2 x + 1 |) + 2 \ln (| x - 1 |) + C$