Calculus Examples

$\int x \ln (1 + x) d x$

Step 1

Integrate by parts using the formula $\int u d v = u v - \int v d u$ , where $u = \ln (1 + x)$ and $d v = x$ .

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

Step 2

Simplify.

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Step 2.1

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

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

Step 2.2

Combine $\ln (1 + x)$ and $\frac{x^{2}}{2}$ .

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

Step 3

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

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

Step 4

Combine $x^{2}$ and $\frac{1}{x + 1}$ .

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

Step 5

Divide $x^{2}$ by $x + 1$ .

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Step 5.1

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$x$

$1$

$x^{2}$

$0 x$

$0$

Step 5.2

Divide the highest order term in the dividend $x^{2}$ by the highest order term in divisor $x$ .

					$x$
	$x$	+	$1$		$x^{2}$	+	$0 x$	+	$0$

Step 5.3

Multiply the new quotient term by the divisor.

				$x$
$x$	+	$1$		$x^{2}$	+	$0 x$	+	$0$
			+	$x^{2}$	+	$x$

Step 5.4

The expression needs to be subtracted from the dividend, so change all the signs in $x^{2} + x$

				$x$
$x$	+	$1$		$x^{2}$	+	$0 x$	+	$0$
			-	$x^{2}$	-	$x$

Step 5.5

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$x$
$x$	+	$1$		$x^{2}$	+	$0 x$	+	$0$
			-	$x^{2}$	-	$x$
					-	$x$

Step 5.6

Pull the next terms from the original dividend down into the current dividend.

				$x$
$x$	+	$1$		$x^{2}$	+	$0 x$	+	$0$
			-	$x^{2}$	-	$x$
					-	$x$	+	$0$

Step 5.7

Divide the highest order term in the dividend $- x$ by the highest order term in divisor $x$ .

				$x$	-	$1$
$x$	+	$1$		$x^{2}$	+	$0 x$	+	$0$
			-	$x^{2}$	-	$x$
					-	$x$	+	$0$

Step 5.8

Multiply the new quotient term by the divisor.

				$x$	-	$1$
$x$	+	$1$		$x^{2}$	+	$0 x$	+	$0$
			-	$x^{2}$	-	$x$
					-	$x$	+	$0$
					-	$x$	-	$1$

Step 5.9

The expression needs to be subtracted from the dividend, so change all the signs in $- x - 1$

				$x$	-	$1$
$x$	+	$1$		$x^{2}$	+	$0 x$	+	$0$
			-	$x^{2}$	-	$x$
					-	$x$	+	$0$
					+	$x$	+	$1$

Step 5.10

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$x$	-	$1$
$x$	+	$1$		$x^{2}$	+	$0 x$	+	$0$
			-	$x^{2}$	-	$x$
					-	$x$	+	$0$
					+	$x$	+	$1$
							+	$1$

Step 5.11

The final answer is the quotient plus the remainder over the divisor.

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

Step 6

Split the single integral into multiple integrals.

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

Step 7

By the Power Rule, the integral of $x$ with respect to $x$ is $\frac{1}{2} x^{2}$ .

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

Step 8

Apply the constant rule.

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

Step 9

Let $u = x + 1$ . Then $d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 9.1

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

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Step 9.1.1

Differentiate $x + 1$ .

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

Step 9.1.2

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 9.1.3

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 9.1.4

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

$1 + 0$

Step 9.1.5

Add $1$ and $0$ .

$1$

Step 9.2

Rewrite the problem using $u$ and $d u$ .

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

Step 10

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

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

Step 11

Simplify.

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Step 11.1

Simplify.

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

Step 11.2

Simplify.

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Step 11.2.1

Combine $\frac{1}{2}$ and $\ln (1 + x)$ .

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

Step 11.2.2

Combine $\frac{\ln (1 + x)}{2}$ and $x^{2}$ .

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

Step 11.2.3

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

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

Step 11.2.4

Combine $- \frac{1}{2} (\frac{1}{2} x^{2} - x + \ln (| u |))$ and $\frac{2}{2}$ .

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

Step 11.2.5

Combine the numerators over the common denominator.

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

Step 11.2.6

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

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

Step 11.2.7

Multiply $2$ by $- 1$ .

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

Step 11.2.8

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

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

Step 11.2.9

Cancel the common factor of $- 2$ and $2$ .

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Step 11.2.9.1

Factor $2$ out of $- 2$ .

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

Step 11.2.9.2

Cancel the common factors.

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Step 11.2.9.2.1

Factor $2$ out of $2$ .

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

Step 11.2.9.2.2

Cancel the common factor.

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

Step 11.2.9.2.3

Rewrite the expression.

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

Step 11.2.9.2.4

Divide $- 1$ by $1$ .

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

Step 12

Replace all occurrences of $u$ with $x + 1$ .

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

Step 13

Simplify.

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Step 13.1

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

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

Step 13.2

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

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

Step 13.3

Combine the numerators over the common denominator.

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

Step 13.4

Move $2$ to the left of $\ln (| x + 1 |)$ .

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

Step 13.5

Apply the distributive property.

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

Step 13.6

Multiply $- - x$ .

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Step 13.6.1

Multiply $- 1$ by $- 1$ .

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

Step 13.6.2

Multiply $x$ by $1$ .

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

Step 14

Reorder terms.

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