Calculus Examples

$\frac{d u}{d t} = \frac{(u + 1) (t + 1)}{(u + 2) (t - 1)}$

Step 1

Separate the variables.

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Step 1.1

Regroup factors.

$\frac{d u}{d t} = \frac{u + 1}{u + 2} \cdot \frac{t + 1}{t - 1}$

Step 1.2

Multiply both sides by $\frac{u + 2}{u + 1}$ .

$\frac{u + 2}{u + 1} \frac{d u}{d t} = \frac{u + 2}{u + 1} (\frac{u + 1}{u + 2} \cdot \frac{t + 1}{t - 1})$

Step 1.3

Simplify.

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Step 1.3.1

Multiply $\frac{u + 1}{u + 2}$ by $\frac{t + 1}{t - 1}$ .

$\frac{u + 2}{u + 1} \frac{d u}{d t} = \frac{u + 2}{u + 1} \cdot \frac{(u + 1) (t + 1)}{(u + 2) (t - 1)}$

Step 1.3.2

Cancel the common factor of $u + 2$ .

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Step 1.3.2.1

Cancel the common factor.

$\frac{u + 2}{u + 1} \frac{d u}{d t} = \frac{u + 2}{u + 1} \cdot \frac{(u + 1) (t + 1)}{(u + 2) (t - 1)}$

Step 1.3.2.2

Rewrite the expression.

$\frac{u + 2}{u + 1} \frac{d u}{d t} = \frac{1}{u + 1} \cdot \frac{(u + 1) (t + 1)}{t - 1}$

Step 1.3.3

Cancel the common factor of $u + 1$ .

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Step 1.3.3.1

Cancel the common factor.

$\frac{u + 2}{u + 1} \frac{d u}{d t} = \frac{1}{u + 1} \cdot \frac{(u + 1) (t + 1)}{t - 1}$

Step 1.3.3.2

Rewrite the expression.

$\frac{u + 2}{u + 1} \frac{d u}{d t} = \frac{t + 1}{t - 1}$

Step 1.4

Rewrite the equation.

$\frac{u + 2}{u + 1} d u = \frac{t + 1}{t - 1} d t$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int \frac{u + 2}{u + 1} d u = \int \frac{t + 1}{t - 1} d t$

Step 2.2

Integrate the left side.

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Step 2.2.1

Divide $u + 2$ by $u + 1$ .

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Step 2.2.1.1

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

$u$

$1$

$u$

$2$

Step 2.2.1.2

Divide the highest order term in the dividend $u$ by the highest order term in divisor $u$ .

					$1$
	$u$	+	$1$		$u$	+	$2$

Step 2.2.1.3

Multiply the new quotient term by the divisor.

				$1$
$u$	+	$1$		$u$	+	$2$
			+	$u$	+	$1$

Step 2.2.1.4

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

				$1$
$u$	+	$1$		$u$	+	$2$
			-	$u$	-	$1$

Step 2.2.1.5

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

				$1$
$u$	+	$1$		$u$	+	$2$
			-	$u$	-	$1$
					+	$1$

Step 2.2.1.6

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

$\int 1 + \frac{1}{u + 1} d u = \int \frac{t + 1}{t - 1} d t$

Step 2.2.2

Split the single integral into multiple integrals.

$\int d u + \int \frac{1}{u + 1} d u = \int \frac{t + 1}{t - 1} d t$

Step 2.2.3

Apply the constant rule.

$u + C_{1} + \int \frac{1}{u + 1} d u = \int \frac{t + 1}{t - 1} d t$

Step 2.2.4

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

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Step 2.2.4.1

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

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Step 2.2.4.1.1

Differentiate $u + 1$ .

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

Step 2.2.4.1.2

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

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

Step 2.2.4.1.3

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

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

Step 2.2.4.1.4

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

$1 + 0$

Step 2.2.4.1.5

Add $1$ and $0$ .

$1$

Step 2.2.4.2

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

$u + C_{1} + \int \frac{1}{u_{1}} d u_{1} = \int \frac{t + 1}{t - 1} d t$

Step 2.2.5

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

$u + C_{1} + \ln (| u_{1} |) + C_{2} = \int \frac{t + 1}{t - 1} d t$

Step 2.2.6

Simplify.

$u + \ln (| u_{1} |) + C_{3} = \int \frac{t + 1}{t - 1} d t$

Step 2.2.7

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

$u + \ln (| u + 1 |) + C_{3} = \int \frac{t + 1}{t - 1} d t$

Step 2.3

Integrate the right side.

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Step 2.3.1

Divide $t + 1$ by $t - 1$ .

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Step 2.3.1.1

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

$t$

$1$

$t$

$1$

Step 2.3.1.2

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

					$1$
	$t$	-	$1$		$t$	+	$1$

Step 2.3.1.3

Multiply the new quotient term by the divisor.

				$1$
$t$	-	$1$		$t$	+	$1$
			+	$t$	-	$1$

Step 2.3.1.4

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

				$1$
$t$	-	$1$		$t$	+	$1$
			-	$t$	+	$1$

Step 2.3.1.5

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

				$1$
$t$	-	$1$		$t$	+	$1$
			-	$t$	+	$1$
					+	$2$

Step 2.3.1.6

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

$u + \ln (| u + 1 |) + C_{3} = \int 1 + \frac{2}{t - 1} d t$

Step 2.3.2

Split the single integral into multiple integrals.

$u + \ln (| u + 1 |) + C_{3} = \int d t + \int \frac{2}{t - 1} d t$

Step 2.3.3

Apply the constant rule.

$u + \ln (| u + 1 |) + C_{3} = t + C_{4} + \int \frac{2}{t - 1} d t$

Step 2.3.4

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

$u + \ln (| u + 1 |) + C_{3} = t + C_{4} + 2 \int \frac{1}{t - 1} d t$

Step 2.3.5

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

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Step 2.3.5.1

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

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Step 2.3.5.1.1

Differentiate $t - 1$ .

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

Step 2.3.5.1.2

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 2.3.5.1.3

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 2.3.5.1.4

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

$1 + 0$

Step 2.3.5.1.5

Add $1$ and $0$ .

$1$

Step 2.3.5.2

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

$u + \ln (| u + 1 |) + C_{3} = t + C_{4} + 2 \int \frac{1}{u_{2}} d u_{2}$

Step 2.3.6

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

$u + \ln (| u + 1 |) + C_{3} = t + C_{4} + 2 (\ln (| u_{2} |) + C_{5})$

Step 2.3.7

Simplify.

$u + \ln (| u + 1 |) + C_{3} = t + 2 \ln (| u_{2} |) + C_{6}$

Step 2.3.8

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

$u + \ln (| u + 1 |) + C_{3} = t + 2 \ln (| t - 1 |) + C_{6}$

Step 2.4

Group the constant of integration on the right side as $C$ .

$u + \ln (| u + 1 |) = t + 2 \ln (| t - 1 |) + C$