Calculus Examples

$\frac{2 x + 3}{2 - x}$

Step 1

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

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

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.

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

Step 4

Reorder $2$ and $- x$ .

$\int \frac{2 x + 3}{- x + 2} d x$

Step 5

Divide $2 x + 3$ by $- x + 2$ .

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Step 5.1

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

$x$

$2$

$2 x$

$3$

Step 5.2

Divide the highest order term in the dividend $2 x$ by the highest order term in divisor $- x$ .

				-	$2$
-	$x$	+	$2$		$2 x$	+	$3$

Step 5.3

Multiply the new quotient term by the divisor.

				-	$2$
-	$x$	+	$2$		$2 x$	+	$3$
				+	$2 x$	-	$4$

Step 5.4

The expression needs to be subtracted from the dividend, so change all the signs in $2 x - 4$

				-	$2$
-	$x$	+	$2$		$2 x$	+	$3$
				-	$2 x$	+	$4$

Step 5.5

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

				-	$2$
-	$x$	+	$2$		$2 x$	+	$3$
				-	$2 x$	+	$4$
						+	$7$

Step 5.6

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

$\int - 2 + \frac{7}{- x + 2} d x$

Step 6

Split the single integral into multiple integrals.

$\int - 2 d x + \int \frac{7}{- x + 2} d x$

Step 7

Apply the constant rule.

$- 2 x + C + \int \frac{7}{- x + 2} d x$

Step 8

Since $7$ is constant with respect to $x$ , move $7$ out of the integral.

$- 2 x + C + 7 \int \frac{1}{- x + 2} d x$

Step 9

Let $u = - x + 2$ . Then $d u = - d x$ , so $- d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 9.1

Let $u = - x + 2$ . Find $\frac{d u}{d x}$ .

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Step 9.1.1

Rewrite.

$\frac{- 1}{1}$

Step 9.1.2

Divide $- 1$ by $1$ .

$- 1$

Step 9.2

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

$- 2 x + C + 7 \int \frac{- 1}{u} d u$

Step 10

Move the negative in front of the fraction.

$- 2 x + C + 7 \int - \frac{1}{u} d u$

Step 11

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

$- 2 x + C + 7 (- \int \frac{1}{u} d u)$

Step 12

Multiply $- 1$ by $7$ .

$- 2 x + C - 7 \int \frac{1}{u} d u$

Step 13

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

$- 2 x + C - 7 (\ln (| u |) + C)$

Step 14

Simplify.

$- 2 x - 7 \ln (| u |) + C$

Step 15

Replace all occurrences of $u$ with $- x + 2$ .

$- 2 x - 7 \ln (| - x + 2 |) + C$

Step 16

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

$F (x) =$ $- 2 x - 7 \ln (| - x + 2 |) + C$