Calculus Examples

$\frac{x^{2} + 2 x + 3}{x + 1}$

Step 1

Write $\frac{x^{2} + 2 x + 3}{x + 1}$ as a function.

$f (x) = \frac{x^{2} + 2 x + 3}{x + 1}$

Step 2

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 3

Set up the integral to solve.

$F (x) = \int \frac{x^{2} + 2 x + 3}{x + 1} d x$

Step 4

Divide $x^{2} + 2 x + 3$ by $x + 1$ .

Tap for more steps...

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 $0$ .

$x$

$1$

$x^{2}$

$2 x$

$3$

Step 4.2

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

					$x$
	$x$	+	$1$		$x^{2}$	+	$2 x$	+	$3$

Step 4.3

Multiply the new quotient term by the divisor.

				$x$
$x$	+	$1$		$x^{2}$	+	$2 x$	+	$3$
			+	$x^{2}$	+	$x$

Step 4.4

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

				$x$
$x$	+	$1$		$x^{2}$	+	$2 x$	+	$3$
			-	$x^{2}$	-	$x$

Step 4.5

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

				$x$
$x$	+	$1$		$x^{2}$	+	$2 x$	+	$3$
			-	$x^{2}$	-	$x$
					+	$x$

Step 4.6

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

				$x$
$x$	+	$1$		$x^{2}$	+	$2 x$	+	$3$
			-	$x^{2}$	-	$x$
					+	$x$	+	$3$

Step 4.7

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

				$x$	+	$1$
$x$	+	$1$		$x^{2}$	+	$2 x$	+	$3$
			-	$x^{2}$	-	$x$
					+	$x$	+	$3$

Step 4.8

Multiply the new quotient term by the divisor.

				$x$	+	$1$
$x$	+	$1$		$x^{2}$	+	$2 x$	+	$3$
			-	$x^{2}$	-	$x$
					+	$x$	+	$3$
					+	$x$	+	$1$

Step 4.9

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

				$x$	+	$1$
$x$	+	$1$		$x^{2}$	+	$2 x$	+	$3$
			-	$x^{2}$	-	$x$
					+	$x$	+	$3$
					-	$x$	-	$1$

Step 4.10

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

				$x$	+	$1$
$x$	+	$1$		$x^{2}$	+	$2 x$	+	$3$
			-	$x^{2}$	-	$x$
					+	$x$	+	$3$
					-	$x$	-	$1$
							+	$2$

Step 4.11

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

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

Step 5

Split the single integral into multiple integrals.

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

Step 6

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

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

Step 7

Apply the constant rule.

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

Step 8

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

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

Step 9

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

Tap for more steps...

Step 9.1

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

Tap for more steps...

Step 9.1.1

Differentiate $x + 1$ .

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

Step 9.1.2

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 9.1.3

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 9.1.4

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

$1 + 0$

Step 9.1.5

Add $1$ and $0$ .

$1$

Step 9.2

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

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

Step 10

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

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

Step 11

Simplify.

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

Step 12

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

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

Step 13

The answer is the antiderivative of the function $f (x) = \frac{x^{2} + 2 x + 3}{x + 1}$ .

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