Calculus Examples

$4 x y d x + (x^{2} + 1) d y = 0$

Step 1

Subtract $4 x y d x$ from both sides of the equation.

$(x^{2} + 1) d y = - 4 x y d x$

Step 2

Multiply both sides by $\frac{1}{(x^{2} + 1) y}$ .

$\frac{1}{(x^{2} + 1) y} (x^{2} + 1) d y = \frac{1}{(x^{2} + 1) y} (- 4 x y) d x$

Step 3

Simplify.

Tap for more steps...

Step 3.1

Cancel the common factor of $x^{2} + 1$ .

Tap for more steps...

Step 3.1.1

Factor $x^{2} + 1$ out of $(x^{2} + 1) y$ .

$\frac{1}{(x^{2} + 1) (y)} (x^{2} + 1) d y = \frac{1}{(x^{2} + 1) y} (- 4 x y) d x$

Step 3.1.2

Cancel the common factor.

$\frac{1}{(x^{2} + 1) y} (x^{2} + 1) d y = \frac{1}{(x^{2} + 1) y} (- 4 x y) d x$

Step 3.1.3

Rewrite the expression.

$\frac{1}{y} d y = \frac{1}{(x^{2} + 1) y} (- 4 x y) d x$

Step 3.2

Rewrite using the commutative property of multiplication.

$\frac{1}{y} d y = - 4 \frac{1}{(x^{2} + 1) y} (x y) d x$

Step 3.3

Combine $- 4$ and $\frac{1}{(x^{2} + 1) y}$ .

$\frac{1}{y} d y = \frac{- 4}{(x^{2} + 1) y} (x y) d x$

Step 3.4

Cancel the common factor of $y$ .

Tap for more steps...

Step 3.4.1

Factor $y$ out of $(x^{2} + 1) y$ .

$\frac{1}{y} d y = \frac{- 4}{y (x^{2} + 1)} (x y) d x$

Step 3.4.2

Factor $y$ out of $x y$ .

$\frac{1}{y} d y = \frac{- 4}{y (x^{2} + 1)} (y x) d x$

Step 3.4.3

Cancel the common factor.

$\frac{1}{y} d y = \frac{- 4}{y (x^{2} + 1)} (y x) d x$

Step 3.4.4

Rewrite the expression.

$\frac{1}{y} d y = \frac{- 4}{x^{2} + 1} x d x$

Step 3.5

Combine $\frac{- 4}{x^{2} + 1}$ and $x$ .

$\frac{1}{y} d y = \frac{- 4 x}{x^{2} + 1} d x$

Step 3.6

Move the negative in front of the fraction.

$\frac{1}{y} d y = - \frac{4 x}{x^{2} + 1} d x$

Step 4

Integrate both sides.

Tap for more steps...

Step 4.1

Set up an integral on each side.

$\int \frac{1}{y} d y = \int - \frac{4 x}{x^{2} + 1} d x$

Step 4.2

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

$\ln (| y |) + C_{1} = \int - \frac{4 x}{x^{2} + 1} d x$

Step 4.3

Integrate the right side.

Tap for more steps...

Step 4.3.1

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

$\ln (| y |) + C_{1} = - \int \frac{4 x}{x^{2} + 1} d x$

Step 4.3.2

Since $4$ is constant with respect to $x$ , move $4$ out of the integral.

$\ln (| y |) + C_{1} = - (4 \int \frac{x}{x^{2} + 1} d x)$

Step 4.3.3

Multiply $4$ by $- 1$ .

$\ln (| y |) + C_{1} = - 4 \int \frac{x}{x^{2} + 1} d x$

Step 4.3.4

Let $u = x^{2} + 1$ . Then $d u = 2 x d x$ , so $\frac{1}{2} d u = x d x$ . Rewrite using $u$ and $d$ $u$ .

Tap for more steps...

Step 4.3.4.1

Let $u = x^{2} + 1$ . Find $\frac{d u}{d x}$ .

Tap for more steps...

Step 4.3.4.1.1

Differentiate $x^{2} + 1$ .

$\frac{d}{d x} [x^{2} + 1]$

Step 4.3.4.1.2

By the Sum Rule, the derivative of $x^{2} + 1$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [1]$ .

$\frac{d}{d x} [x^{2}] + \frac{d}{d x} [1]$

Step 4.3.4.1.3

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$2 x + \frac{d}{d x} [1]$

Step 4.3.4.1.4

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

$2 x + 0$

Step 4.3.4.1.5

Add $2 x$ and $0$ .

$2 x$

Step 4.3.4.2

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

$\ln (| y |) + C_{1} = - 4 \int \frac{1}{u} \cdot \frac{1}{2} d u$

Step 4.3.5

Simplify.

Tap for more steps...

Step 4.3.5.1

Multiply $\frac{1}{u}$ by $\frac{1}{2}$ .

$\ln (| y |) + C_{1} = - 4 \int \frac{1}{u \cdot 2} d u$

Step 4.3.5.2

Move $2$ to the left of $u$ .

$\ln (| y |) + C_{1} = - 4 \int \frac{1}{2 u} d u$

Step 4.3.6

Since $\frac{1}{2}$ is constant with respect to $u$ , move $\frac{1}{2}$ out of the integral.

$\ln (| y |) + C_{1} = - 4 (\frac{1}{2} \int \frac{1}{u} d u)$

Step 4.3.7

Simplify.

Tap for more steps...

Step 4.3.7.1

Combine $\frac{1}{2}$ and $- 4$ .

$\ln (| y |) + C_{1} = \frac{- 4}{2} \int \frac{1}{u} d u$

Step 4.3.7.2

Cancel the common factor of $- 4$ and $2$ .

Tap for more steps...

Step 4.3.7.2.1

Factor $2$ out of $- 4$ .

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

Step 4.3.7.2.2

Cancel the common factors.

Tap for more steps...

Step 4.3.7.2.2.1

Factor $2$ out of $2$ .

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

Step 4.3.7.2.2.2

Cancel the common factor.

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

Step 4.3.7.2.2.3

Rewrite the expression.

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

Step 4.3.7.2.2.4

Divide $- 2$ by $1$ .

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

Step 4.3.8

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

$\ln (| y |) + C_{1} = - 2 (\ln (| u |) + C_{2})$

Step 4.3.9

Simplify.

$\ln (| y |) + C_{1} = - 2 \ln (| u |) + C_{2}$

Step 4.3.10

Replace all occurrences of $u$ with $x^{2} + 1$ .

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

Step 4.4

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

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

Step 5

Solve for $y$ .

Tap for more steps...

Step 5.1

Move all the terms containing a logarithm to the left side of the equation.

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

Step 5.2

Simplify the left side.

Tap for more steps...

Step 5.2.1

Simplify $\ln (| y |) + 2 \ln (| x^{2} + 1 |)$ .

Tap for more steps...

Step 5.2.1.1

Simplify each term.

Tap for more steps...

Step 5.2.1.1.1

Simplify $2 \ln (| x^{2} + 1 |)$ by moving $2$ inside the logarithm.

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

Step 5.2.1.1.2

Remove the absolute value in ${| x^{2} + 1 |}^{2}$ because exponentiations with even powers are always positive.

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

Step 5.2.1.2

Use the product property of logarithms, $\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

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

Step 5.3

To solve for $y$ , rewrite the equation using properties of logarithms.

$e^{\ln (| y | {(x^{2} + 1)}^{2})} = e^{C}$

Step 5.4

Rewrite $\ln (| y | {(x^{2} + 1)}^{2}) = C$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b \neq 1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$e^{C} = | y | {(x^{2} + 1)}^{2}$

Step 5.5

Solve for $y$ .

Tap for more steps...

Step 5.5.1

Rewrite the equation as $| y | {(x^{2} + 1)}^{2} = e^{C}$ .

$| y | {(x^{2} + 1)}^{2} = e^{C}$

Step 5.5.2

Divide each term in $| y | {(x^{2} + 1)}^{2} = e^{C}$ by ${(x^{2} + 1)}^{2}$ and simplify.

Tap for more steps...

Step 5.5.2.1

Divide each term in $| y | {(x^{2} + 1)}^{2} = e^{C}$ by ${(x^{2} + 1)}^{2}$ .

$\frac{| y | {(x^{2} + 1)}^{2}}{{(x^{2} + 1)}^{2}} = \frac{e^{C}}{{(x^{2} + 1)}^{2}}$

Step 5.5.2.2

Simplify the left side.

Tap for more steps...

Step 5.5.2.2.1

Cancel the common factor of ${(x^{2} + 1)}^{2}$ .

Tap for more steps...

Step 5.5.2.2.1.1

Cancel the common factor.

$\frac{| y | {(x^{2} + 1)}^{2}}{{(x^{2} + 1)}^{2}} = \frac{e^{C}}{{(x^{2} + 1)}^{2}}$

Step 5.5.2.2.1.2

Divide $| y |$ by $1$ .

$| y | = \frac{e^{C}}{{(x^{2} + 1)}^{2}}$

Step 5.5.3

Remove the absolute value term. This creates a $\pm$ on the right side of the equation because $| x | = \pm x$ .

$y = \pm \frac{e^{C}}{{(x^{2} + 1)}^{2}}$

Step 6

Group the constant terms together.

Tap for more steps...

Step 6.1

Simplify the constant of integration.

$y = \pm \frac{C}{{(x^{2} + 1)}^{2}}$

Step 6.2

Combine constants with the plus or minus.

$y = C \frac{1}{{(x^{2} + 1)}^{2}}$