Calculus Examples

$\frac{y - 1}{- y - 1} \cdot \frac{d y}{d x} = \frac{1}{1 + x^{2}}$

Step 1

Rewrite the equation.

$\frac{y - 1}{- y - 1} d y = \frac{1}{1 + x^{2}} d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

Integrate the left side.

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Step 2.2.1

Divide $y - 1$ by $- y - 1$ .

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Step 2.2.1.1

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

$y$

$1$

$y$

$1$

Step 2.2.1.2

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

				-	$1$
-	$y$	-	$1$		$y$	-	$1$

Step 2.2.1.3

Multiply the new quotient term by the divisor.

				-	$1$
-	$y$	-	$1$		$y$	-	$1$
				+	$y$	+	$1$

Step 2.2.1.4

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

				-	$1$
-	$y$	-	$1$		$y$	-	$1$
				-	$y$	-	$1$

Step 2.2.1.5

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

				-	$1$
-	$y$	-	$1$		$y$	-	$1$
				-	$y$	-	$1$
						-	$2$

Step 2.2.1.6

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

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

Step 2.2.2

Split the single integral into multiple integrals.

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

Step 2.2.3

Apply the constant rule.

$- y + C_{1} + \int - \frac{2}{- y - 1} d y = \int \frac{1}{1 + x^{2}} d x$

Step 2.2.4

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

$- y + C_{1} - \int \frac{2}{- y - 1} d y = \int \frac{1}{1 + x^{2}} d x$

Step 2.2.5

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

$- y + C_{1} - (2 \int \frac{1}{- y - 1} d y) = \int \frac{1}{1 + x^{2}} d x$

Step 2.2.6

Multiply $2$ by $- 1$ .

$- y + C_{1} - 2 \int \frac{1}{- y - 1} d y = \int \frac{1}{1 + x^{2}} d x$

Step 2.2.7

Let $u = - y - 1$ . Then $d u = - d y$ , so $- d u = d y$ . Rewrite using $u$ and $d$ $u$ .

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Step 2.2.7.1

Let $u = - y - 1$ . Find $\frac{d u}{d y}$ .

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Step 2.2.7.1.1

Rewrite.

$\frac{- 1}{1}$

Step 2.2.7.1.2

Divide $- 1$ by $1$ .

$- 1$

Step 2.2.7.2

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

$- y + C_{1} - 2 \int \frac{- 1}{u} d u = \int \frac{1}{1 + x^{2}} d x$

Step 2.2.8

Move the negative in front of the fraction.

$- y + C_{1} - 2 \int - \frac{1}{u} d u = \int \frac{1}{1 + x^{2}} d x$

Step 2.2.9

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

$- y + C_{1} - 2 (- \int \frac{1}{u} d u) = \int \frac{1}{1 + x^{2}} d x$

Step 2.2.10

Multiply $- 1$ by $- 2$ .

$- y + C_{1} + 2 \int \frac{1}{u} d u = \int \frac{1}{1 + x^{2}} d x$

Step 2.2.11

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

$- y + C_{1} + 2 (\ln (| u |) + C_{2}) = \int \frac{1}{1 + x^{2}} d x$

Step 2.2.12

Simplify.

$- y + 2 \ln (| u |) + C_{3} = \int \frac{1}{1 + x^{2}} d x$

Step 2.2.13

Replace all occurrences of $u$ with $- y - 1$ .

$- y + 2 \ln (| - y - 1 |) + C_{3} = \int \frac{1}{1 + x^{2}} d x$

Step 2.3

Integrate the right side.

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Step 2.3.1

Rewrite $1$ as $1^{2}$ .

$- y + 2 \ln (| - y - 1 |) + C_{3} = \int \frac{1}{1^{2} + x^{2}} d x$

Step 2.3.2

The integral of $\frac{1}{1^{2} + x^{2}}$ with respect to $x$ is $\arctan (x) + C_{4}$ .

$- y + 2 \ln (| - y - 1 |) + C_{3} = \arctan (x) + C_{4}$

Step 2.4

Group the constant of integration on the right side as $C$ .

$- y + 2 \ln (| - y - 1 |) = \arctan (x) + C$