Calculus Examples

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

Step 1

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

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

Step 2

Simplify.

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Step 2.1

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

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

Step 2.2

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

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

Step 3

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

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

Step 4

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

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

Step 5

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

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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 $0$ .

$t$

$1$

$t^{2}$

$0 t$

$0$

Step 5.2

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

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

Step 5.3

Multiply the new quotient term by the divisor.

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

Step 5.4

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

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

Step 5.5

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

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

Step 5.6

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

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

Step 5.7

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

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

Step 5.8

Multiply the new quotient term by the divisor.

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

Step 5.9

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

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

Step 5.10

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

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

Step 5.11

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

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

Step 6

Split the single integral into multiple integrals.

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

Step 7

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

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

Step 8

Apply the constant rule.

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

Step 9

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

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Step 9.1

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

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Step 9.1.1

Differentiate $t + 1$ .

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

Step 9.1.2

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

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

Step 9.1.3

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

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

Step 9.1.4

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

$1 + 0$

Step 9.1.5

Add $1$ and $0$ .

$1$

Step 9.2

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

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

Step 10

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

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

Step 11

Simplify.

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Step 11.1

Simplify.

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

Step 11.2

Simplify.

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Step 11.2.1

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

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

Step 11.2.2

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

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

Step 11.2.3

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

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

Step 11.2.4

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

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

Step 11.2.5

Combine the numerators over the common denominator.

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

Step 11.2.6

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

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

Step 11.2.7

Multiply $2$ by $- 1$ .

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

Step 11.2.8

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

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

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 $- 2$ .

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

Step 11.2.9.2

Cancel the common factors.

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Step 11.2.9.2.1

Factor $2$ out of $2$ .

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

Step 11.2.9.2.2

Cancel the common factor.

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

Step 11.2.9.2.3

Rewrite the expression.

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

Step 11.2.9.2.4

Divide $- 1$ by $1$ .

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

Step 12

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

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

Step 13

Simplify.

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Step 13.1

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

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

Step 13.2

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

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

Step 13.3

Combine the numerators over the common denominator.

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

Step 13.4

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

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

Step 13.5

Apply the distributive property.

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

Step 13.6

Multiply $- - t$ .

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Step 13.6.1

Multiply $- 1$ by $- 1$ .

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

Step 13.6.2

Multiply $t$ by $1$ .

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

Step 14

Reorder terms.

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