Calculus Examples

$\int \frac{y - 2}{y^{2} + 2 y - 3} d y$

Step 1

Write the fraction using partial fraction decomposition.

Tap for more steps...

Step 1.1

Decompose the fraction and multiply through by the common denominator.

Tap for more steps...

Step 1.1.1

Factor $y^{2} + 2 y - 3$ using the AC method.

Tap for more steps...

Step 1.1.1.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 3$ and whose sum is $2$ .

$- 1, 3$

Step 1.1.1.2

Write the factored form using these integers.

$\frac{y - 2}{(y - 1) (y + 3)}$

Step 1.1.2

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 $A$ .

$\frac{A}{y - 1}$

Step 1.1.3

$\frac{A}{y - 1} + \frac{B}{y + 3}$

Step 1.1.4

Multiply each fraction in the equation by the denominator of the original expression. In this case, the denominator is $(y - 1) (y + 3)$ .

$\frac{(y - 2) (y - 1) (y + 3)}{(y - 1) (y + 3)} = \frac{(A) (y - 1) (y + 3)}{y - 1} + \frac{(B) (y - 1) (y + 3)}{y + 3}$

Step 1.1.5

Cancel the common factor of $y - 1$ .

Tap for more steps...

Step 1.1.5.1

Cancel the common factor.

$\frac{(y - 2) (y - 1) (y + 3)}{(y - 1) (y + 3)} = \frac{(A) (y - 1) (y + 3)}{y - 1} + \frac{(B) (y - 1) (y + 3)}{y + 3}$

Step 1.1.5.2

Rewrite the expression.

$\frac{(y - 2) (y + 3)}{y + 3} = \frac{(A) (y - 1) (y + 3)}{y - 1} + \frac{(B) (y - 1) (y + 3)}{y + 3}$

Step 1.1.6

Cancel the common factor of $y + 3$ .

Tap for more steps...

Step 1.1.6.1

Cancel the common factor.

$\frac{(y - 2) (y + 3)}{y + 3} = \frac{(A) (y - 1) (y + 3)}{y - 1} + \frac{(B) (y - 1) (y + 3)}{y + 3}$

Step 1.1.6.2

Divide $y - 2$ by $1$ .

$y - 2 = \frac{(A) (y - 1) (y + 3)}{y - 1} + \frac{(B) (y - 1) (y + 3)}{y + 3}$

Step 1.1.7

Simplify each term.

Tap for more steps...

Step 1.1.7.1

Cancel the common factor of $y - 1$ .

Tap for more steps...

Step 1.1.7.1.1

Cancel the common factor.

$y - 2 = \frac{A (y - 1) (y + 3)}{y - 1} + \frac{(B) (y - 1) (y + 3)}{y + 3}$

Step 1.1.7.1.2

Divide $(A) (y + 3)$ by $1$ .

$y - 2 = (A) (y + 3) + \frac{(B) (y - 1) (y + 3)}{y + 3}$

Step 1.1.7.2

Apply the distributive property.

$y - 2 = A y + A \cdot 3 + \frac{(B) (y - 1) (y + 3)}{y + 3}$

Step 1.1.7.3

Move $3$ to the left of $A$ .

$y - 2 = A y + 3 \cdot A + \frac{(B) (y - 1) (y + 3)}{y + 3}$

Step 1.1.7.4

Cancel the common factor of $y + 3$ .

Tap for more steps...

Step 1.1.7.4.1

Cancel the common factor.

$y - 2 = A y + 3 A + \frac{(B) (y - 1) (y + 3)}{y + 3}$

Step 1.1.7.4.2

Divide $(B) (y - 1)$ by $1$ .

$y - 2 = A y + 3 A + (B) (y - 1)$

Step 1.1.7.5

Apply the distributive property.

$y - 2 = A y + 3 A + B y + B \cdot - 1$

Step 1.1.7.6

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

$y - 2 = A y + 3 A + B y - 1 \cdot B$

Step 1.1.7.7

Rewrite $- 1 B$ as $- B$ .

$y - 2 = A y + 3 A + B y - B$

Step 1.1.8

Move $3 A$ .

$y - 2 = A y + B y + 3 A - B$

Step 1.2

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

Tap for more steps...

Step 1.2.1

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

$1 = A + B$

Step 1.2.2

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

$- 2 = 3 A - 1 B$

Step 1.2.3

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

$1 = A + B$

$- 2 = 3 A - 1 B$

$1 = A + B$

$- 2 = 3 A - 1 B$

Step 1.3

Solve the system of equations.

Tap for more steps...

Step 1.3.1

Solve for $A$ in $1 = A + B$ .

Tap for more steps...

Step 1.3.1.1

Rewrite the equation as $A + B = 1$ .

$A + B = 1$

$- 2 = 3 A - 1 B$

Step 1.3.1.2

Subtract $B$ from both sides of the equation.

$A = 1 - B$

$- 2 = 3 A - 1 B$

$A = 1 - B$

$- 2 = 3 A - 1 B$

Step 1.3.2

Replace all occurrences of $A$ with $1 - B$ in each equation.

Tap for more steps...

Step 1.3.2.1

Replace all occurrences of $A$ in $- 2 = 3 A - 1 B$ with $1 - B$ .

$- 2 = 3 (1 - B) - 1 B$

$A = 1 - B$

Step 1.3.2.2

Simplify the right side.

Tap for more steps...

Step 1.3.2.2.1

Simplify $3 (1 - B) - 1 B$ .

Tap for more steps...

Step 1.3.2.2.1.1

Simplify each term.

Tap for more steps...

Step 1.3.2.2.1.1.1

Apply the distributive property.

$- 2 = 3 \cdot 1 + 3 (- B) - 1 B$

$A = 1 - B$

Step 1.3.2.2.1.1.2

Multiply $3$ by $1$ .

$- 2 = 3 + 3 (- B) - 1 B$

$A = 1 - B$

Step 1.3.2.2.1.1.3

Multiply $- 1$ by $3$ .

$- 2 = 3 - 3 B - 1 B$

$A = 1 - B$

Step 1.3.2.2.1.1.4

Rewrite $- 1 B$ as $- B$ .

$- 2 = 3 - 3 B - B$

$A = 1 - B$

$- 2 = 3 - 3 B - B$

$A = 1 - B$

Step 1.3.2.2.1.2

Subtract $B$ from $- 3 B$ .

$- 2 = 3 - 4 B$

$A = 1 - B$

$- 2 = 3 - 4 B$

$A = 1 - B$

$- 2 = 3 - 4 B$

$A = 1 - B$

$- 2 = 3 - 4 B$

$A = 1 - B$

Step 1.3.3

Solve for $B$ in $- 2 = 3 - 4 B$ .

Tap for more steps...

Step 1.3.3.1

Rewrite the equation as $3 - 4 B = - 2$ .

$3 - 4 B = - 2$

$A = 1 - B$

Step 1.3.3.2

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

Tap for more steps...

Step 1.3.3.2.1

Subtract $3$ from both sides of the equation.

$- 4 B = - 2 - 3$

$A = 1 - B$

Step 1.3.3.2.2

Subtract $3$ from $- 2$ .

$- 4 B = - 5$

$A = 1 - B$

$- 4 B = - 5$

$A = 1 - B$

Step 1.3.3.3

Divide each term in $- 4 B = - 5$ by $- 4$ and simplify.

Tap for more steps...

Step 1.3.3.3.1

Divide each term in $- 4 B = - 5$ by $- 4$ .

$\frac{- 4 B}{- 4} = \frac{- 5}{- 4}$

$A = 1 - B$

Step 1.3.3.3.2

Simplify the left side.

Tap for more steps...

Step 1.3.3.3.2.1

Cancel the common factor of $- 4$ .

Tap for more steps...

Step 1.3.3.3.2.1.1

Cancel the common factor.

$\frac{- 4 B}{- 4} = \frac{- 5}{- 4}$

$A = 1 - B$

Step 1.3.3.3.2.1.2

Divide $B$ by $1$ .

$B = \frac{- 5}{- 4}$

$A = 1 - B$

$B = \frac{- 5}{- 4}$

$A = 1 - B$

$B = \frac{- 5}{- 4}$

$A = 1 - B$

Step 1.3.3.3.3

Simplify the right side.

Tap for more steps...

Step 1.3.3.3.3.1

Dividing two negative values results in a positive value.

$B = \frac{5}{4}$

$A = 1 - B$

$B = \frac{5}{4}$

$A = 1 - B$

$B = \frac{5}{4}$

$A = 1 - B$

$B = \frac{5}{4}$

$A = 1 - B$

Step 1.3.4

Replace all occurrences of $B$ with $\frac{5}{4}$ in each equation.

Tap for more steps...

Step 1.3.4.1

Replace all occurrences of $B$ in $A = 1 - B$ with $\frac{5}{4}$ .

$A = 1 - (\frac{5}{4})$

$B = \frac{5}{4}$

Step 1.3.4.2

Simplify the right side.

Tap for more steps...

Step 1.3.4.2.1

Simplify $1 - (\frac{5}{4})$ .

Tap for more steps...

Step 1.3.4.2.1.1

Write $1$ as a fraction with a common denominator.

$A = \frac{4}{4} - \frac{5}{4}$

$B = \frac{5}{4}$

Step 1.3.4.2.1.2

Combine the numerators over the common denominator.

$A = \frac{4 - 5}{4}$

$B = \frac{5}{4}$

Step 1.3.4.2.1.3

Subtract $5$ from $4$ .

$A = \frac{- 1}{4}$

$B = \frac{5}{4}$

Step 1.3.4.2.1.4

Move the negative in front of the fraction.

$A = - \frac{1}{4}$

$B = \frac{5}{4}$

$A = - \frac{1}{4}$

$B = \frac{5}{4}$

$A = - \frac{1}{4}$

$B = \frac{5}{4}$

$A = - \frac{1}{4}$

$B = \frac{5}{4}$

Step 1.3.5

List all of the solutions.

$A = - \frac{1}{4}, B = \frac{5}{4}$

Step 1.4

Replace each of the partial fraction coefficients in $\frac{A}{y - 1} + \frac{B}{y + 3}$ with the values found for $A$ and $B$ .

$\frac{- \frac{1}{4}}{y - 1} + \frac{\frac{5}{4}}{y + 3}$

Step 1.5

Simplify.

Tap for more steps...

Step 1.5.1

Multiply the numerator by the reciprocal of the denominator.

$- \frac{1}{4} \cdot \frac{1}{y - 1} + \frac{\frac{5}{4}}{y + 3}$

Step 1.5.2

Multiply $\frac{1}{y - 1}$ by $\frac{1}{4}$ .

$- \frac{1}{(y - 1) \cdot 4} + \frac{\frac{5}{4}}{y + 3}$

Step 1.5.3

Move $4$ to the left of $y - 1$ .

$- \frac{1}{4 (y - 1)} + \frac{\frac{5}{4}}{y + 3}$

Step 1.5.4

Multiply the numerator by the reciprocal of the denominator.

$- \frac{1}{4 (y - 1)} + \frac{5}{4} \cdot \frac{1}{y + 3}$

Step 1.5.5

Multiply $\frac{5}{4}$ by $\frac{1}{y + 3}$ .

$\int - \frac{1}{4 (y - 1)} + \frac{5}{4 (y + 3)} d y$

Step 2

Split the single integral into multiple integrals.

$\int - \frac{1}{4 (y - 1)} d y + \int \frac{5}{4 (y + 3)} d y$

Step 3

Since $- 1$ is constant with respect to $y$ , move $- 1$ out of the integral.

$- \int \frac{1}{4 (y - 1)} d y + \int \frac{5}{4 (y + 3)} d y$

Step 4

Since $\frac{1}{4}$ is constant with respect to $y$ , move $\frac{1}{4}$ out of the integral.

$- (\frac{1}{4} \int \frac{1}{y - 1} d y) + \int \frac{5}{4 (y + 3)} d y$

Step 5

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

Tap for more steps...

Step 5.1

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

Tap for more steps...

Step 5.1.1

Differentiate $y - 1$ .

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

Step 5.1.2

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

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

Step 5.1.3

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

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

Step 5.1.4

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

$1 + 0$

Step 5.1.5

Add $1$ and $0$ .

$1$

Step 5.2

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

$- \frac{1}{4} \int \frac{1}{u_{1}} d u_{1} + \int \frac{5}{4 (y + 3)} d y$

Step 6

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

$- \frac{1}{4} (\ln (| u_{1} |) + C) + \int \frac{5}{4 (y + 3)} d y$

Step 7

Since $\frac{5}{4}$ is constant with respect to $y$ , move $\frac{5}{4}$ out of the integral.

$- \frac{1}{4} (\ln (| u_{1} |) + C) + \frac{5}{4} \int \frac{1}{y + 3} d y$

Step 8

Let $u_{2} = y + 3$ . Then $d u_{2} = d y$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

Tap for more steps...

Step 8.1

Let $u_{2} = y + 3$ . Find $\frac{d u_{2}}{d y}$ .

Tap for more steps...

Step 8.1.1

Differentiate $y + 3$ .

$\frac{d}{d y} [y + 3]$

Step 8.1.2

By the Sum Rule, the derivative of $y + 3$ with respect to $y$ is $\frac{d}{d y} [y] + \frac{d}{d y} [3]$ .

$\frac{d}{d y} [y] + \frac{d}{d y} [3]$

Step 8.1.3

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

$1 + \frac{d}{d y} [3]$

Step 8.1.4

Since $3$ is constant with respect to $y$ , the derivative of $3$ with respect to $y$ is $0$ .

$1 + 0$

Step 8.1.5

Add $1$ and $0$ .

$1$

Step 8.2

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

$- \frac{1}{4} (\ln (| u_{1} |) + C) + \frac{5}{4} \int \frac{1}{u_{2}} d u_{2}$

Step 9

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

$- \frac{1}{4} (\ln (| u_{1} |) + C) + \frac{5}{4} (\ln (| u_{2} |) + C)$

Step 10

Simplify.

$- \frac{1}{4} \ln (| u_{1} |) + \frac{5}{4} \ln (| u_{2} |) + C$

Step 11

Substitute back in for each integration substitution variable.

Tap for more steps...

Step 11.1

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

$- \frac{1}{4} \ln (| y - 1 |) + \frac{5}{4} \ln (| u_{2} |) + C$

Step 11.2

Replace all occurrences of $u_{2}$ with $y + 3$ .

$- \frac{1}{4} \ln (| y - 1 |) + \frac{5}{4} \ln (| y + 3 |) + C$