Calculus Examples

$\frac{225 + 10 x}{1 + 0.05 x}$

Step 1

Let $u = 1 + 0.05 x$ . Then $d u = 0.05 d x$ , so $\frac{1}{0.05} d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 1.1

Let $u = 1 + 0.05 x$ . Find $\frac{d u}{d x}$ .

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Step 1.1.1

Rewrite.

$\frac{0.05}{1}$

Step 1.1.2

Divide $0.05$ by $1$ .

$0.05$

Step 1.2

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

$\int \frac{4500 + 200 (20 u - 20)}{u} d u$

Step 2

Split the fraction into multiple fractions.

$\int \frac{4500}{u} + \frac{200 (20 u - 20)}{u} d u$

Step 3

Split the single integral into multiple integrals.

$\int \frac{4500}{u} d u + \int \frac{200 (20 u - 20)}{u} d u$

Step 4

Since $4500$ is constant with respect to $u$ , move $4500$ out of the integral.

$4500 \int \frac{1}{u} d u + \int \frac{200 (20 u - 20)}{u} d u$

Step 5

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

$4500 (\ln (| u |) + C) + \int \frac{200 (20 u - 20)}{u} d u$

Step 6

Since $200$ is constant with respect to $u$ , move $200$ out of the integral.

$4500 (\ln (| u |) + C) + 200 \int \frac{20 u - 20}{u} d u$

Step 7

Divide $20 u - 20$ by $u$ .

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Step 7.1

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

$u$

$0$

$20 u$

$20$

Step 7.2

Divide the highest order term in the dividend $20 u$ by the highest order term in divisor $u$ .

					$20$
	$u$	+	$0$		$20 u$	-	$20$

Step 7.3

Multiply the new quotient term by the divisor.

				$20$
$u$	+	$0$		$20 u$	-	$20$
			+	$20 u$	+	$0$

Step 7.4

The expression needs to be subtracted from the dividend, so change all the signs in $20 u + 0$

				$20$
$u$	+	$0$		$20 u$	-	$20$
			-	$20 u$	-	$0$

Step 7.5

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

				$20$
$u$	+	$0$		$20 u$	-	$20$
			-	$20 u$	-	$0$
					-	$20$

Step 7.6

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

$4500 (\ln (| u |) + C) + 200 \int 20 - \frac{20}{u} d u$

Step 8

Split the single integral into multiple integrals.

$4500 (\ln (| u |) + C) + 200 (\int 20 d u + \int - \frac{20}{u} d u)$

Step 9

Apply the constant rule.

$4500 (\ln (| u |) + C) + 200 (20 u + C + \int - \frac{20}{u} d u)$

Step 10

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

$4500 (\ln (| u |) + C) + 200 (20 u + C - \int \frac{20}{u} d u)$

Step 11

Since $20$ is constant with respect to $u$ , move $20$ out of the integral.

$4500 (\ln (| u |) + C) + 200 (20 u + C - (20 \int \frac{1}{u} d u))$

Step 12

Multiply $20$ by $- 1$ .

$4500 (\ln (| u |) + C) + 200 (20 u + C - 20 \int \frac{1}{u} d u)$

Step 13

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

$4500 (\ln (| u |) + C) + 200 (20 u + C - 20 (\ln (| u |) + C))$

Step 14

Simplify.

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Step 14.1

Simplify.

$4500 \ln (| u |) + 4000 u - 4000 \ln (| u |) + C$

Step 14.2

Subtract $4000 \ln (| u |)$ from $4500 \ln (| u |)$ .

$4000 u + 500 \ln (| u |) + C$

Step 15

Replace all occurrences of $u$ with $1 + 0.05 x$ .

$4000 (1 + 0.05 x) + 500 \ln (| 1 + 0.05 x |) + C$