Calculus Examples

$3 (y^{4} + 1) d x + 4 x y^{3} d y = 0$

Step 1

Subtract $3 (y^{4} + 1) d x$ from both sides of the equation.

$4 x y^{3} d y = - 3 (y^{4} + 1) d x$

Step 2

Multiply both sides by $\frac{1}{x (y^{4} + 1)}$ .

$\frac{1}{x (y^{4} + 1)} (4 x y^{3}) d y = \frac{1}{x (y^{4} + 1)} (- 3 (y^{4} + 1)) d x$

Step 3

Simplify.

Tap for more steps...

Step 3.1

Rewrite using the commutative property of multiplication.

$4 \frac{1}{x (y^{4} + 1)} (x y^{3}) d y = \frac{1}{x (y^{4} + 1)} (- 3 (y^{4} + 1)) d x$

Step 3.2

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

$\frac{4}{x (y^{4} + 1)} (x y^{3}) d y = \frac{1}{x (y^{4} + 1)} (- 3 (y^{4} + 1)) d x$

Step 3.3

Cancel the common factor of $x$ .

Tap for more steps...

Step 3.3.1

Factor $x$ out of $x y^{3}$ .

$\frac{4}{x (y^{4} + 1)} (x (y^{3})) d y = \frac{1}{x (y^{4} + 1)} (- 3 (y^{4} + 1)) d x$

Step 3.3.2

Cancel the common factor.

$\frac{4}{x (y^{4} + 1)} (x y^{3}) d y = \frac{1}{x (y^{4} + 1)} (- 3 (y^{4} + 1)) d x$

Step 3.3.3

Rewrite the expression.

$\frac{4}{y^{4} + 1} y^{3} d y = \frac{1}{x (y^{4} + 1)} (- 3 (y^{4} + 1)) d x$

Step 3.4

Combine $\frac{4}{y^{4} + 1}$ and $y^{3}$ .

$\frac{4 y^{3}}{y^{4} + 1} d y = \frac{1}{x (y^{4} + 1)} (- 3 (y^{4} + 1)) d x$

Step 3.5

Rewrite using the commutative property of multiplication.

$\frac{4 y^{3}}{y^{4} + 1} d y = - 3 \frac{1}{x (y^{4} + 1)} (y^{4} + 1) d x$

Step 3.6

Combine $- 3$ and $\frac{1}{x (y^{4} + 1)}$ .

$\frac{4 y^{3}}{y^{4} + 1} d y = \frac{- 3}{x (y^{4} + 1)} (y^{4} + 1) d x$

Step 3.7

Cancel the common factor of $y^{4} + 1$ .

Tap for more steps...

Step 3.7.1

Factor $y^{4} + 1$ out of $x (y^{4} + 1)$ .

$\frac{4 y^{3}}{y^{4} + 1} d y = \frac{- 3}{(y^{4} + 1) x} (y^{4} + 1) d x$

Step 3.7.2

Cancel the common factor.

$\frac{4 y^{3}}{y^{4} + 1} d y = \frac{- 3}{(y^{4} + 1) x} (y^{4} + 1) d x$

Step 3.7.3

Rewrite the expression.

$\frac{4 y^{3}}{y^{4} + 1} d y = \frac{- 3}{x} d x$

Step 3.8

Move the negative in front of the fraction.

$\frac{4 y^{3}}{y^{4} + 1} d y = - \frac{3}{x} d x$

Step 4

Integrate both sides.

Tap for more steps...

Step 4.1

Set up an integral on each side.

$\int \frac{4 y^{3}}{y^{4} + 1} d y = \int - \frac{3}{x} d x$

Step 4.2

Integrate the left side.

Tap for more steps...

Step 4.2.1

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

$4 \int \frac{y^{3}}{y^{4} + 1} d y = \int - \frac{3}{x} d x$

Step 4.2.2

Rewrite $y^{4}$ as ${(y^{4})}^{1}$ .

$4 \int \frac{y^{3}}{{(y^{4})}^{1} + 1} d y = \int - \frac{3}{x} d x$

Step 4.2.3

Let $u_{1} = y^{4}$ . Then $d u_{1} = 4 y^{3} d y$ , so $\frac{1}{4} d u_{1} = y^{3} d y$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

Tap for more steps...

Step 4.2.3.1

Let $u_{1} = y^{4}$ . Find $\frac{d u_{1}}{d y}$ .

Tap for more steps...

Step 4.2.3.1.1

Differentiate $y^{4}$ .

$\frac{d}{d y} [y^{4}]$

Step 4.2.3.1.2

Differentiate using the Power Rule which states that $\frac{d}{d y} [y^{n}]$ is $n y^{n - 1}$ where $n = 4$ .

$4 y^{3}$

Step 4.2.3.2

Rewrite the problem using $u_{1}$ and $d u_{1}$ .

$4 \int \frac{1}{{u_{1}}^{1} + 1} \cdot \frac{1}{4} d u_{1} = \int - \frac{3}{x} d x$

Step 4.2.4

Simplify.

Tap for more steps...

Step 4.2.4.1

Simplify.

$4 \int \frac{1}{u_{1} + 1} \cdot \frac{1}{4} d u_{1} = \int - \frac{3}{x} d x$

Step 4.2.4.2

Multiply $\frac{1}{u_{1} + 1}$ by $\frac{1}{4}$ .

$4 \int \frac{1}{(u_{1} + 1) \cdot 4} d u_{1} = \int - \frac{3}{x} d x$

Step 4.2.4.3

Move $4$ to the left of $u_{1} + 1$ .

$4 \int \frac{1}{4 (u_{1} + 1)} d u_{1} = \int - \frac{3}{x} d x$

Step 4.2.5

Since $\frac{1}{4}$ is constant with respect to $u_{1}$ , move $\frac{1}{4}$ out of the integral.

$4 (\frac{1}{4} \int \frac{1}{u_{1} + 1} d u_{1}) = \int - \frac{3}{x} d x$

Step 4.2.6

Simplify.

Tap for more steps...

Step 4.2.6.1

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

$\frac{4}{4} \int \frac{1}{u_{1} + 1} d u_{1} = \int - \frac{3}{x} d x$

Step 4.2.6.2

Cancel the common factor of $4$ .

Tap for more steps...

Step 4.2.6.2.1

Cancel the common factor.

$\frac{4}{4} \int \frac{1}{u_{1} + 1} d u_{1} = \int - \frac{3}{x} d x$

Step 4.2.6.2.2

Rewrite the expression.

$1 \int \frac{1}{u_{1} + 1} d u_{1} = \int - \frac{3}{x} d x$

Step 4.2.6.3

Multiply $\int \frac{1}{u_{1} + 1} d u_{1}$ by $1$ .

$\int \frac{1}{u_{1} + 1} d u_{1} = \int - \frac{3}{x} d x$

Step 4.2.7

Let $u_{2} = u_{1} + 1$ . Then $d u_{2} = d u_{1}$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

Tap for more steps...

Step 4.2.7.1

Let $u_{2} = u_{1} + 1$ . Find $\frac{d u_{2}}{d u_{1}}$ .

Tap for more steps...

Step 4.2.7.1.1

Differentiate $u_{1} + 1$ .

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

Step 4.2.7.1.2

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

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

Step 4.2.7.1.3

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

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

Step 4.2.7.1.4

Since $1$ is constant with respect to $u_{1}$ , the derivative of $1$ with respect to $u_{1}$ is $0$ .

$1 + 0$

Step 4.2.7.1.5

Add $1$ and $0$ .

$1$

Step 4.2.7.2

Rewrite the problem using $u_{2}$ and $d u_{2}$ .

$\int \frac{1}{u_{2}} d u_{2} = \int - \frac{3}{x} d x$

Step 4.2.8

The integral of $\frac{1}{u_{2}}$ with respect to $u_{2}$ is $\ln (| u_{2} |)$ .

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

Step 4.2.9

Substitute back in for each integration substitution variable.

Tap for more steps...

Step 4.2.9.1

Replace all occurrences of $u_{2}$ with $u_{1} + 1$ .

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

Step 4.2.9.2

Replace all occurrences of $u_{1}$ with $y^{4}$ .

$\ln (| y^{4} + 1 |) + C_{1} = \int - \frac{3}{x} d x$

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.

$\ln (| y^{4} + 1 |) + C_{1} = - \int \frac{3}{x} d x$

Step 4.3.2

Since $3$ is constant with respect to $x$ , move $3$ out of the integral.

$\ln (| y^{4} + 1 |) + C_{1} = - (3 \int \frac{1}{x} d x)$

Step 4.3.3

Multiply $3$ by $- 1$ .

$\ln (| y^{4} + 1 |) + C_{1} = - 3 \int \frac{1}{x} d x$

Step 4.3.4

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

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

Step 4.3.5

Simplify.

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

Step 4.4

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

$\ln (| y^{4} + 1 |) = - 3 \ln (| x |) + C$

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.

$\ln (| y^{4} + 1 |) + 3 \ln (| x |) = C$

Step 5.2

Simplify the left side.

Tap for more steps...

Step 5.2.1

Simplify $\ln (| y^{4} + 1 |) + 3 \ln (| x |)$ .

Tap for more steps...

Step 5.2.1.1

Simplify $3 \ln (| x |)$ by moving $3$ inside the logarithm.

$\ln (| y^{4} + 1 |) + \ln ({| x |}^{3}) = C$

Step 5.2.1.2

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

$\ln (| y^{4} + 1 | {| x |}^{3}) = C$

Step 5.3

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

$e^{\ln (| y^{4} + 1 | {| x |}^{3})} = e^{C}$

Step 5.4

Rewrite $\ln (| y^{4} + 1 | {| x |}^{3}) = C$ 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 $b^{y} = x$ .

$e^{C} = | y^{4} + 1 | {| x |}^{3}$

Step 5.5

Solve for $y$ .

Tap for more steps...

Step 5.5.1

Rewrite the equation as $| y^{4} + 1 | {| x |}^{3} = e^{C}$ .

$| y^{4} + 1 | {| x |}^{3} = e^{C}$

Step 5.5.2

Divide each term in $| y^{4} + 1 | {| x |}^{3} = e^{C}$ by ${| x |}^{3}$ and simplify.

Tap for more steps...

Step 5.5.2.1

Divide each term in $| y^{4} + 1 | {| x |}^{3} = e^{C}$ by ${| x |}^{3}$ .

$\frac{| y^{4} + 1 | {| x |}^{3}}{{| x |}^{3}} = \frac{e^{C}}{{| x |}^{3}}$

Step 5.5.2.2

Simplify the left side.

Tap for more steps...

Step 5.5.2.2.1

Cancel the common factor of ${| x |}^{3}$ .

Tap for more steps...

Step 5.5.2.2.1.1

Cancel the common factor.

$\frac{| y^{4} + 1 | {| x |}^{3}}{{| x |}^{3}} = \frac{e^{C}}{{| x |}^{3}}$

Step 5.5.2.2.1.2

Divide $| y^{4} + 1 |$ by $1$ .

$| y^{4} + 1 | = \frac{e^{C}}{{| x |}^{3}}$

Step 5.5.3

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

$y^{4} + 1 = \pm \frac{e^{C}}{{| x |}^{3}}$

Step 5.5.4

Subtract $1$ from both sides of the equation.

$y^{4} = \pm \frac{e^{C}}{{| x |}^{3}} - 1$

Step 5.5.5

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$y = \pm \sqrt[4]{\pm \frac{e^{C}}{{| x |}^{3}} - 1}$

Step 6

Group the constant terms together.

Tap for more steps...

Step 6.1

Simplify the constant of integration.

$y = \pm \sqrt[4]{\pm \frac{C}{{| x |}^{3}} - 1}$

Step 6.2

Combine constants with the plus or minus.

$y = \pm \sqrt[4]{C \frac{1}{{| x |}^{3}} - 1}$