Calculus Examples

$\int_{0}^{1} \frac{1}{1 - x^{2}} d x$

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

Factor the fraction.

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Step 1.1.1.1

Rewrite $1$ as $1^{2}$ .

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

Step 1.1.1.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 1$ and $b = x$ .

$\frac{1}{(1 + x) (1 - x)}$

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 $A$ .

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

Step 1.1.3

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

Step 1.1.4

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

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

Step 1.1.5

Cancel the common factor of $1 + x$ .

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Step 1.1.5.1

Cancel the common factor.

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

Step 1.1.5.2

Rewrite the expression.

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

Step 1.1.6

Cancel the common factor of $1 - x$ .

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Step 1.1.6.1

Cancel the common factor.

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

Step 1.1.6.2

Rewrite the expression.

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

Step 1.1.7

Simplify each term.

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Step 1.1.7.1

Cancel the common factor of $1 + x$ .

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Step 1.1.7.1.1

Cancel the common factor.

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

Step 1.1.7.1.2

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

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

Step 1.1.7.2

Apply the distributive property.

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

Step 1.1.7.3

Multiply $A$ by $1$ .

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

Step 1.1.7.4

Rewrite using the commutative property of multiplication.

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

Step 1.1.7.5

Cancel the common factor of $1 - x$ .

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Step 1.1.7.5.1

Cancel the common factor.

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

Step 1.1.7.5.2

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

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

Step 1.1.7.6

Apply the distributive property.

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

Step 1.1.7.7

Multiply $B$ by $1$ .

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

Step 1.1.8

Simplify the expression.

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Step 1.1.8.1

Move $A$ .

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

Step 1.1.8.2

Reorder $B$ and $x$ .

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

Step 1.1.8.3

Move $B$ .

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

Step 1.1.8.4

Move $A$ .

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

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.

$0 = - 1 A + B$

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.

$1 = A + B$

Step 1.2.3

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

$0 = - 1 A + B$

$1 = A + B$

$0 = - 1 A + B$

$1 = A + B$

Step 1.3

Solve the system of equations.

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Step 1.3.1

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

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Step 1.3.1.1

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

$A + B = 1$

$0 = - 1 A + B$

Step 1.3.1.2

Subtract $B$ from both sides of the equation.

$A = 1 - B$

$0 = - 1 A + B$

$A = 1 - B$

$0 = - 1 A + B$

Step 1.3.2

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

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Step 1.3.2.1

Replace all occurrences of $A$ in $0 = - 1 A + B$ with $1 - B$ .

$0 = - 1 (1 - B) + B$

$A = 1 - B$

Step 1.3.2.2

Simplify the right side.

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Step 1.3.2.2.1

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

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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.

$0 = - 1 \cdot 1 - 1 (- B) + B$

$A = 1 - B$

Step 1.3.2.2.1.1.2

Multiply $- 1$ by $1$ .

$0 = - 1 - 1 (- B) + B$

$A = 1 - B$

Step 1.3.2.2.1.1.3

Multiply $- 1 (- B)$ .

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Step 1.3.2.2.1.1.3.1

Multiply $- 1$ by $- 1$ .

$0 = - 1 + 1 B + B$

$A = 1 - B$

Step 1.3.2.2.1.1.3.2

Multiply $B$ by $1$ .

$0 = - 1 + B + B$

$A = 1 - B$

$0 = - 1 + B + B$

$A = 1 - B$

$0 = - 1 + B + B$

$A = 1 - B$

Step 1.3.2.2.1.2

Add $B$ and $B$ .

$0 = - 1 + 2 B$

$A = 1 - B$

$0 = - 1 + 2 B$

$A = 1 - B$

$0 = - 1 + 2 B$

$A = 1 - B$

$0 = - 1 + 2 B$

$A = 1 - B$

Step 1.3.3

Solve for $B$ in $0 = - 1 + 2 B$ .

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Step 1.3.3.1

Rewrite the equation as $- 1 + 2 B = 0$ .

$- 1 + 2 B = 0$

$A = 1 - B$

Step 1.3.3.2

Add $1$ to both sides of the equation.

$2 B = 1$

$A = 1 - B$

Step 1.3.3.3

Divide each term in $2 B = 1$ by $2$ and simplify.

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Step 1.3.3.3.1

Divide each term in $2 B = 1$ by $2$ .

$\frac{2 B}{2} = \frac{1}{2}$

$A = 1 - B$

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 $2$ .

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Step 1.3.3.3.2.1.1

Cancel the common factor.

$\frac{2 B}{2} = \frac{1}{2}$

$A = 1 - B$

Step 1.3.3.3.2.1.2

Divide $B$ by $1$ .

$B = \frac{1}{2}$

$A = 1 - B$

$B = \frac{1}{2}$

$A = 1 - B$

$B = \frac{1}{2}$

$A = 1 - B$

$B = \frac{1}{2}$

$A = 1 - B$

$B = \frac{1}{2}$

$A = 1 - B$

Step 1.3.4

Replace all occurrences of $B$ with $\frac{1}{2}$ in each equation.

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Step 1.3.4.1

Replace all occurrences of $B$ in $A = 1 - B$ with $\frac{1}{2}$ .

$A = 1 - (\frac{1}{2})$

$B = \frac{1}{2}$

Step 1.3.4.2

Simplify the right side.

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Step 1.3.4.2.1

Simplify $1 - (\frac{1}{2})$ .

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Step 1.3.4.2.1.1

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

$A = \frac{2}{2} - \frac{1}{2}$

$B = \frac{1}{2}$

Step 1.3.4.2.1.2

Combine the numerators over the common denominator.

$A = \frac{2 - 1}{2}$

$B = \frac{1}{2}$

Step 1.3.4.2.1.3

Subtract $1$ from $2$ .

$A = \frac{1}{2}$

$B = \frac{1}{2}$

$A = \frac{1}{2}$

$B = \frac{1}{2}$

$A = \frac{1}{2}$

$B = \frac{1}{2}$

$A = \frac{1}{2}$

$B = \frac{1}{2}$

Step 1.3.5

List all of the solutions.

$A = \frac{1}{2}, B = \frac{1}{2}$

Step 1.4

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

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

Step 1.5

Simplify.

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Step 1.5.1

Multiply the numerator by the reciprocal of the denominator.

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

Step 1.5.2

Multiply $\frac{1}{2}$ by $\frac{1}{1 + x}$ .

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

Step 1.5.3

Multiply the numerator by the reciprocal of the denominator.

$\frac{1}{2 (1 + x)} + \frac{1}{2} \cdot \frac{1}{1 - x}$

Step 1.5.4

Multiply $\frac{1}{2}$ by $\frac{1}{1 - x}$ .

$\int_{0}^{1} \frac{1}{2 (1 + x)} + \frac{1}{2 (1 - x)} d x$

Step 2

Split the single integral into multiple integrals.

$\int_{0}^{1} \frac{1}{2 (1 + x)} d x + \int_{0}^{1} \frac{1}{2 (1 - x)} d x$

Step 3

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

$\frac{1}{2} \int_{0}^{1} \frac{1}{1 + x} d x + \int_{0}^{1} \frac{1}{2 (1 - x)} d x$

Step 4

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

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Step 4.1

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

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Step 4.1.1

Differentiate $1 + x$ .

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

Step 4.1.2

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

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

Step 4.1.3

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

$0 + \frac{d}{d x} [x]$

Step 4.1.4

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

$0 + 1$

Step 4.1.5

Add $0$ and $1$ .

$1$

Step 4.2

Substitute the lower limit in for $x$ in $u_{1} = 1 + x$ .

$u_{lower} = 1 + 0$

Step 4.3

Add $1$ and $0$ .

$u_{lower} = 1$

Step 4.4

Substitute the upper limit in for $x$ in $u_{1} = 1 + x$ .

$u_{upper} = 1 + 1$

Step 4.5

Add $1$ and $1$ .

$u_{upper} = 2$

Step 4.6

The values found for $u_{lower}$ and $u_{upper}$ will be used to evaluate the definite integral.

$u_{lower} = 1$

$u_{upper} = 2$

Step 4.7

Rewrite the problem using $u_{1}$ , $d u_{1}$ , and the new limits of integration.

$\frac{1}{2} \int_{1}^{2} \frac{1}{u_{1}} d u_{1} + \int_{0}^{1} \frac{1}{2 (1 - x)} d x$

Step 5

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

$\frac{1}{2} \ln (| u_{1} |)]_{1}^{2} + \int_{0}^{1} \frac{1}{2 (1 - x)} d x$

Step 6

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

$\frac{1}{2} \ln (| u_{1} |)]_{1}^{2} + \frac{1}{2} \int_{0}^{1} \frac{1}{1 - x} d x$

Step 7

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

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Step 7.1

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

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Step 7.1.1

Rewrite.

$\frac{- 1}{1}$

Step 7.1.2

Divide $- 1$ by $1$ .

$- 1$

Step 7.2

Substitute the lower limit in for $x$ in $u_{2} = 1 - x$ .

$u_{lower} = 1 - 0$

Step 7.3

Subtract $0$ from $1$ .

$u_{lower} = 1$

Step 7.4

Substitute the upper limit in for $x$ in $u_{2} = 1 - x$ .

$u_{upper} = 1 - 1 \cdot 1$

Step 7.5

Simplify.

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Step 7.5.1

Multiply $- 1$ by $1$ .

$u_{upper} = 1 - 1$

Step 7.5.2

Subtract $1$ from $1$ .

$u_{upper} = 0$

Step 7.6

The values found for $u_{lower}$ and $u_{upper}$ will be used to evaluate the definite integral.

$u_{lower} = 1$

$u_{upper} = 0$

Step 7.7

Rewrite the problem using $u_{2}$ , $d u_{2}$ , and the new limits of integration.

$\frac{1}{2} \ln (| u_{1} |)]_{1}^{2} + \frac{1}{2} \int_{1}^{0} \frac{- 1}{u_{2}} d u_{2}$

Step 8

Move the negative in front of the fraction.

$\frac{1}{2} \ln (| u_{1} |)]_{1}^{2} + \frac{1}{2} \int_{1}^{0} - \frac{1}{u_{2}} d u_{2}$

Step 9

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

$\frac{1}{2} \ln (| u_{1} |)]_{1}^{2} + \frac{1}{2} (- \int_{1}^{0} \frac{1}{u_{2}} d u_{2})$

Step 10

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

$\frac{1}{2} \ln (| u_{1} |)]_{1}^{2} + \frac{1}{2} (- (\ln (| u_{2} |)]_{1}^{0}))$

Step 11

Combine $\frac{1}{2}$ and $\ln (| u_{2} |)]_{1}^{0}$ .

$\frac{1}{2} \ln (| u_{1} |)]_{1}^{2} - \frac{\ln (| u_{2} |)]_{1}^{0}}{2}$

Step 12

Substitute and simplify.

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Step 12.1

Evaluate $\ln (| u_{1} |)$ at $2$ and at $1$ .

$\frac{1}{2} ((\ln (| 2 |)) - \ln (| 1 |)) - \frac{\ln (| u_{2} |)]_{1}^{0}}{2}$

Step 12.2

Evaluate $\ln (| u_{2} |)$ at $0$ and at $1$ .

$\frac{1}{2} ((\ln (| 2 |)) - \ln (| 1 |)) - \frac{(\ln (| 0 |)) - \ln (| 1 |)}{2}$

Step 12.3

Simplify.

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Step 12.3.1

To write $\frac{1}{2} (\ln (| 2 |) - \ln (| 1 |))$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\frac{1}{2} (\ln (| 2 |) - \ln (| 1 |)) \cdot \frac{2}{2} - \frac{\ln (| 0 |) - \ln (| 1 |)}{2}$

Step 12.3.2

Combine $\frac{1}{2} (\ln (| 2 |) - \ln (| 1 |))$ and $\frac{2}{2}$ .

$\frac{\frac{1}{2} (\ln (| 2 |) - \ln (| 1 |)) \cdot 2}{2} - \frac{\ln (| 0 |) - \ln (| 1 |)}{2}$

Step 12.3.3

Combine the numerators over the common denominator.

$\frac{\frac{1}{2} (\ln (| 2 |) - \ln (| 1 |)) \cdot 2 - (\ln (| 0 |) - \ln (| 1 |))}{2}$

Step 12.3.4

Combine $2$ and $\frac{1}{2}$ .

$\frac{\frac{2}{2} (\ln (| 2 |) - \ln (| 1 |)) - (\ln (| 0 |) - \ln (| 1 |))}{2}$

Step 12.3.5

Cancel the common factor of $2$ .

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Step 12.3.5.1

Cancel the common factor.

$\frac{\frac{2}{2} (\ln (| 2 |) - \ln (| 1 |)) - (\ln (| 0 |) - \ln (| 1 |))}{2}$

Step 12.3.5.2

Rewrite the expression.

$\frac{1 (\ln (| 2 |) - \ln (| 1 |)) - (\ln (| 0 |) - \ln (| 1 |))}{2}$

Step 12.3.6

Multiply $\ln (| 2 |) - \ln (| 1 |)$ by $1$ .

$\frac{\ln (| 2 |) - \ln (| 1 |) - (\ln (| 0 |) - \ln (| 1 |))}{2}$

Step 13

Simplify.

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Step 13.1

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\frac{\ln (\frac{| 2 |}{| 1 |}) - (\ln (| 0 |) - \ln (| 1 |))}{2}$

Step 13.2

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\frac{\ln (\frac{| 2 |}{| 1 |}) - \ln (\frac{| 0 |}{| 1 |})}{2}$

Step 13.3

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\frac{\ln (\frac{\frac{| 2 |}{| 1 |}}{\frac{| 0 |}{| 1 |}})}{2}$

Step 13.4

Rewrite $\frac{\frac{| 2 |}{| 1 |}}{\frac{| 0 |}{| 1 |}}$ as a product.

$\frac{\ln (\frac{| 2 |}{| 1 |} \cdot \frac{1}{\frac{| 0 |}{| 1 |}})}{2}$

Step 13.5

Multiply by the reciprocal of the fraction to divide by $\frac{| 0 |}{| 1 |}$ .

$\frac{\ln (\frac{| 2 |}{| 1 |} (1 \frac{| 1 |}{| 0 |}))}{2}$

Step 13.6

Multiply $\frac{| 1 |}{| 0 |}$ by $1$ .

$\frac{\ln (\frac{| 2 |}{| 1 |} \cdot \frac{| 1 |}{| 0 |})}{2}$

Step 13.7

Multiply $\frac{| 2 |}{| 1 |}$ by $\frac{| 1 |}{| 0 |}$ .

$\frac{\ln (\frac{| 2 | | 1 |}{| 1 | | 0 |})}{2}$

Step 13.8

To multiply absolute values, multiply the terms inside each absolute value.

$\frac{\ln (\frac{| 2 \cdot 1 |}{| 1 | | 0 |})}{2}$

Step 13.9

Multiply $2$ by $1$ .

$\frac{\ln (\frac{| 2 |}{| 1 | | 0 |})}{2}$

Step 13.10

To multiply absolute values, multiply the terms inside each absolute value.

$\frac{\ln (\frac{| 2 |}{| 1 \cdot 0 |})}{2}$

Step 13.11

Multiply $0$ by $1$ .

$\frac{\ln (\frac{| 2 |}{| 0 |})}{2}$

Step 14

Simplify.

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Step 14.1

The absolute value is the distance between a number and zero. The distance between $0$ and $0$ is $0$ .

$\frac{\ln (\frac{| 2 |}{0})}{2}$

Step 14.2

The expression contains a division by $0$ . The expression is undefined.

Undefined

$\frac{\ln (\frac{| 2 |}{0})}{2}$

Step 15

The expression contains a division by $0$ . The expression is undefined.

Undefined