Calculus Examples

$f (x) = \frac{6 x^{5} - 17 x^{4} + 9 x^{3} + 10 x^{2}}{x^{3}}$

Step 1

The function $F (x)$ can be found by finding the indefinite integral of the derivative $f (x)$ .

$F (x) = \int f (x) d x$

Step 2

Set up the integral to solve.

$F (x) = \int \frac{6 x^{5} - 17 x^{4} + 9 x^{3} + 10 x^{2}}{x^{3}} d x$

Step 3

Simplify.

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Step 3.1

Factor $x^{2}$ out of $6 x^{5} - 17 x^{4} + 9 x^{3} + 10 x^{2}$ .

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Step 3.1.1

Factor $x^{2}$ out of $6 x^{5}$ .

$\int \frac{x^{2} (6 x^{3}) - 17 x^{4} + 9 x^{3} + 10 x^{2}}{x^{3}} d x$

Step 3.1.2

Factor $x^{2}$ out of $- 17 x^{4}$ .

$\int \frac{x^{2} (6 x^{3}) + x^{2} (- 17 x^{2}) + 9 x^{3} + 10 x^{2}}{x^{3}} d x$

Step 3.1.3

Factor $x^{2}$ out of $9 x^{3}$ .

$\int \frac{x^{2} (6 x^{3}) + x^{2} (- 17 x^{2}) + x^{2} (9 x) + 10 x^{2}}{x^{3}} d x$

Step 3.1.4

Factor $x^{2}$ out of $10 x^{2}$ .

$\int \frac{x^{2} (6 x^{3}) + x^{2} (- 17 x^{2}) + x^{2} (9 x) + x^{2} \cdot 10}{x^{3}} d x$

Step 3.1.5

Factor $x^{2}$ out of $x^{2} (6 x^{3}) + x^{2} (- 17 x^{2})$ .

$\int \frac{x^{2} (6 x^{3} - 17 x^{2}) + x^{2} (9 x) + x^{2} \cdot 10}{x^{3}} d x$

Step 3.1.6

Factor $x^{2}$ out of $x^{2} (6 x^{3} - 17 x^{2}) + x^{2} (9 x)$ .

$\int \frac{x^{2} (6 x^{3} - 17 x^{2} + 9 x) + x^{2} \cdot 10}{x^{3}} d x$

Step 3.1.7

Factor $x^{2}$ out of $x^{2} (6 x^{3} - 17 x^{2} + 9 x) + x^{2} \cdot 10$ .

$\int \frac{x^{2} (6 x^{3} - 17 x^{2} + 9 x + 10)}{x^{3}} d x$

Step 3.2

Cancel the common factors.

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Step 3.2.1

Factor $x^{2}$ out of $x^{3}$ .

$\int \frac{x^{2} (6 x^{3} - 17 x^{2} + 9 x + 10)}{x^{2} x} d x$

Step 3.2.2

Cancel the common factor.

$\int \frac{x^{2} (6 x^{3} - 17 x^{2} + 9 x + 10)}{x^{2} x} d x$

Step 3.2.3

Rewrite the expression.

$\int \frac{6 x^{3} - 17 x^{2} + 9 x + 10}{x} d x$

Step 4

Divide $6 x^{3} - 17 x^{2} + 9 x + 10$ by $x$ .

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Step 4.1

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

$x$

$0$

$6 x^{3}$

$17 x^{2}$

$9 x$

$10$

Step 4.2

Divide the highest order term in the dividend $6 x^{3}$ by the highest order term in divisor $x$ .

					$6 x^{2}$
	$x$	+	$0$		$6 x^{3}$	-	$17 x^{2}$	+	$9 x$	+	$10$

Step 4.3

Multiply the new quotient term by the divisor.

				$6 x^{2}$
$x$	+	$0$		$6 x^{3}$	-	$17 x^{2}$	+	$9 x$	+	$10$
			+	$6 x^{3}$	+	$0$

Step 4.4

The expression needs to be subtracted from the dividend, so change all the signs in $6 x^{3} + 0$

				$6 x^{2}$
$x$	+	$0$		$6 x^{3}$	-	$17 x^{2}$	+	$9 x$	+	$10$
			-	$6 x^{3}$	-	$0$

Step 4.5

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

				$6 x^{2}$
$x$	+	$0$		$6 x^{3}$	-	$17 x^{2}$	+	$9 x$	+	$10$
			-	$6 x^{3}$	-	$0$
					-	$17 x^{2}$

Step 4.6

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

				$6 x^{2}$
$x$	+	$0$		$6 x^{3}$	-	$17 x^{2}$	+	$9 x$	+	$10$
			-	$6 x^{3}$	-	$0$
					-	$17 x^{2}$	+	$9 x$

Step 4.7

Divide the highest order term in the dividend $- 17 x^{2}$ by the highest order term in divisor $x$ .

				$6 x^{2}$	-	$17 x$
$x$	+	$0$		$6 x^{3}$	-	$17 x^{2}$	+	$9 x$	+	$10$
			-	$6 x^{3}$	-	$0$
					-	$17 x^{2}$	+	$9 x$

Step 4.8

Multiply the new quotient term by the divisor.

				$6 x^{2}$	-	$17 x$
$x$	+	$0$		$6 x^{3}$	-	$17 x^{2}$	+	$9 x$	+	$10$
			-	$6 x^{3}$	-	$0$
					-	$17 x^{2}$	+	$9 x$
					-	$17 x^{2}$	+	$0$

Step 4.9

The expression needs to be subtracted from the dividend, so change all the signs in $- 17 x^{2} + 0$

				$6 x^{2}$	-	$17 x$
$x$	+	$0$		$6 x^{3}$	-	$17 x^{2}$	+	$9 x$	+	$10$
			-	$6 x^{3}$	-	$0$
					-	$17 x^{2}$	+	$9 x$
					+	$17 x^{2}$	-	$0$

Step 4.10

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

				$6 x^{2}$	-	$17 x$
$x$	+	$0$		$6 x^{3}$	-	$17 x^{2}$	+	$9 x$	+	$10$
			-	$6 x^{3}$	-	$0$
					-	$17 x^{2}$	+	$9 x$
					+	$17 x^{2}$	-	$0$
							+	$9 x$

Step 4.11

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

				$6 x^{2}$	-	$17 x$
$x$	+	$0$		$6 x^{3}$	-	$17 x^{2}$	+	$9 x$	+	$10$
			-	$6 x^{3}$	-	$0$
					-	$17 x^{2}$	+	$9 x$
					+	$17 x^{2}$	-	$0$
							+	$9 x$	+	$10$

Step 4.12

Divide the highest order term in the dividend $9 x$ by the highest order term in divisor $x$ .

				$6 x^{2}$	-	$17 x$	+	$9$
$x$	+	$0$		$6 x^{3}$	-	$17 x^{2}$	+	$9 x$	+	$10$
			-	$6 x^{3}$	-	$0$
					-	$17 x^{2}$	+	$9 x$
					+	$17 x^{2}$	-	$0$
							+	$9 x$	+	$10$

Step 4.13

Multiply the new quotient term by the divisor.

				$6 x^{2}$	-	$17 x$	+	$9$
$x$	+	$0$		$6 x^{3}$	-	$17 x^{2}$	+	$9 x$	+	$10$
			-	$6 x^{3}$	-	$0$
					-	$17 x^{2}$	+	$9 x$
					+	$17 x^{2}$	-	$0$
							+	$9 x$	+	$10$
							+	$9 x$	+	$0$

Step 4.14

The expression needs to be subtracted from the dividend, so change all the signs in $9 x + 0$

				$6 x^{2}$	-	$17 x$	+	$9$
$x$	+	$0$		$6 x^{3}$	-	$17 x^{2}$	+	$9 x$	+	$10$
			-	$6 x^{3}$	-	$0$
					-	$17 x^{2}$	+	$9 x$
					+	$17 x^{2}$	-	$0$
							+	$9 x$	+	$10$
							-	$9 x$	-	$0$

Step 4.15

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

				$6 x^{2}$	-	$17 x$	+	$9$
$x$	+	$0$		$6 x^{3}$	-	$17 x^{2}$	+	$9 x$	+	$10$
			-	$6 x^{3}$	-	$0$
					-	$17 x^{2}$	+	$9 x$
					+	$17 x^{2}$	-	$0$
							+	$9 x$	+	$10$
							-	$9 x$	-	$0$
									+	$10$

Step 4.16

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

$\int 6 x^{2} - 17 x + 9 + \frac{10}{x} d x$

Step 5

Split the single integral into multiple integrals.

$\int 6 x^{2} d x + \int - 17 x d x + \int 9 d x + \int \frac{10}{x} d x$

Step 6

Since $6$ is constant with respect to $x$ , move $6$ out of the integral.

$6 \int x^{2} d x + \int - 17 x d x + \int 9 d x + \int \frac{10}{x} d x$

Step 7

By the Power Rule, the integral of $x^{2}$ with respect to $x$ is $\frac{1}{3} x^{3}$ .

$6 (\frac{1}{3} x^{3} + C) + \int - 17 x d x + \int 9 d x + \int \frac{10}{x} d x$

Step 8

Since $- 17$ is constant with respect to $x$ , move $- 17$ out of the integral.

$6 (\frac{1}{3} x^{3} + C) - 17 \int x d x + \int 9 d x + \int \frac{10}{x} d x$

Step 9

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

$6 (\frac{1}{3} x^{3} + C) - 17 (\frac{1}{2} x^{2} + C) + \int 9 d x + \int \frac{10}{x} d x$

Step 10

Apply the constant rule.

$6 (\frac{1}{3} x^{3} + C) - 17 (\frac{1}{2} x^{2} + C) + 9 x + C + \int \frac{10}{x} d x$

Step 11

Simplify.

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Step 11.1

Combine $\frac{1}{3}$ and $x^{3}$ .

$6 (\frac{x^{3}}{3} + C) - 17 (\frac{1}{2} x^{2} + C) + 9 x + C + \int \frac{10}{x} d x$

Step 11.2

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

$6 (\frac{x^{3}}{3} + C) - 17 (\frac{x^{2}}{2} + C) + 9 x + C + \int \frac{10}{x} d x$

Step 12

Since $10$ is constant with respect to $x$ , move $10$ out of the integral.

$6 (\frac{x^{3}}{3} + C) - 17 (\frac{x^{2}}{2} + C) + 9 x + C + 10 \int \frac{1}{x} d x$

Step 13

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

$6 (\frac{x^{3}}{3} + C) - 17 (\frac{x^{2}}{2} + C) + 9 x + C + 10 (\ln (| x |) + C)$

Step 14

Simplify.

$2 x^{3} - \frac{17 x^{2}}{2} + 9 x + 10 \ln (| x |) + C$

Step 15

Reorder terms.

$2 x^{3} - \frac{17}{2} x^{2} + 9 x + 10 \ln (| x |) + C$

Step 16

The answer is the antiderivative of the function $f (x) = \frac{6 x^{5} - 17 x^{4} + 9 x^{3} + 10 x^{2}}{x^{3}}$ .

$F (x) =$ $2 x^{3} - \frac{17}{2} x^{2} + 9 x + 10 \ln (| x |) + C$