Calculus Examples

$\int \frac{x^{2}}{x + 1} d x$

Step 1

This integral could not be completed using integration by parts. Mathway will use another method.

Step 2

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

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Step 2.1

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 2.2

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 2.3

Multiply the new quotient term by the divisor.

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

Step 2.4

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 2.5

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 2.6

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 2.7

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 2.8

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 2.9

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 2.10

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 2.11

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

$\int x - 1 + \frac{1}{x + 1} d x$

Step 3

Split the single integral into multiple integrals.

$\int x d x + \int - 1 d x + \int \frac{1}{x + 1} d x$

Step 4

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

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

Step 5

Apply the constant rule.

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

Step 6

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

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Step 6.1

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

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Step 6.1.1

Differentiate $x + 1$ .

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

Step 6.1.2

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 6.1.3

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 6.1.4

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

$1 + 0$

Step 6.1.5

Add $1$ and $0$ .

$1$

Step 6.2

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

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

Step 7

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

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

Step 8

Simplify.

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

Step 9

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

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