Calculus Examples

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

Step 1

Multiply both sides by $\frac{1}{x y}$ .

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

Step 2

Simplify.

Tap for more steps...

Step 2.1

Cancel the common factor of $x$ .

Tap for more steps...

Step 2.1.1

Factor $x$ out of $x y$ .

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

Step 2.1.2

Cancel the common factor.

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

Step 2.1.3

Rewrite the expression.

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

Step 2.2

Cancel the common factor of $y$ .

Tap for more steps...

Step 2.2.1

Factor $y$ out of $x y$ .

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

Step 2.2.2

Cancel the common factor.

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

Step 2.2.3

Rewrite the expression.

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

Step 2.3

Multiply $\frac{1}{x}$ by $x e^{2 x} + 1$ .

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

Step 3

Integrate both sides.

Tap for more steps...

Step 3.1

Set up an integral on each side.

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

Step 3.2

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

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

Step 3.3

Integrate the right side.

Tap for more steps...

Step 3.3.1

Split the fraction into multiple fractions.

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

Step 3.3.2

Split the single integral into multiple integrals.

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

Step 3.3.3

Cancel the common factor of $x$ .

Tap for more steps...

Step 3.3.3.1

Cancel the common factor.

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

Step 3.3.3.2

Divide $e^{2 x}$ by $1$ .

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

Step 3.3.4

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

Tap for more steps...

Step 3.3.4.1

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

Tap for more steps...

Step 3.3.4.1.1

Differentiate $2 x$ .

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

Step 3.3.4.1.2

Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

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

Step 3.3.4.1.3

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

$2 \cdot 1$

Step 3.3.4.1.4

Multiply $2$ by $1$ .

$2$

Step 3.3.4.2

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

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

Step 3.3.5

Combine $e^{u}$ and $\frac{1}{2}$ .

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

Step 3.3.6

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

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

Step 3.3.7

The integral of $e^{u}$ with respect to $u$ is $e^{u}$ .

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

Step 3.3.8

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

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

Step 3.3.9

Simplify.

$\ln (| y |) + C_{1} = \frac{1}{2} e^{u} + \ln (| x |) + C_{4}$

Step 3.3.10

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

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

Step 3.4

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

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

Step 4

Solve for $y$ .

Tap for more steps...

Step 4.1

Simplify the right side.

Tap for more steps...

Step 4.1.1

Combine $\frac{1}{2}$ and $e^{2 x}$ .

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

Step 4.2

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

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

Step 4.3

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

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

Step 4.4

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

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

Step 4.5

Rewrite $\ln (\frac{| y |}{| x |}) = \frac{e^{2 x}}{2} + C$ 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 $b^{y} = x$ .

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

Step 4.6

Solve for $y$ .

Tap for more steps...

Step 4.6.1

Rewrite the equation as $\frac{| y |}{| x |} = e^{\frac{e^{2 x}}{2} + C}$ .

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

Step 4.6.2

Multiply both sides by $| x |$ .

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

Step 4.6.3

Simplify the left side.

Tap for more steps...

Step 4.6.3.1

Cancel the common factor of $| x |$ .

Tap for more steps...

Step 4.6.3.1.1

Cancel the common factor.

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

Step 4.6.3.1.2

Rewrite the expression.

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

Step 4.6.4

Solve for $y$ .

Tap for more steps...

Step 4.6.4.1

Reorder factors in $e^{\frac{e^{2 x}}{2} + C} | x |$ .

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

Step 4.6.4.2

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

$y = \pm | x | e^{\frac{e^{2 x}}{2} + C}$

Step 5

Group the constant terms together.

Tap for more steps...

Step 5.1

Rewrite $e^{\frac{e^{2 x}}{2} + C}$ as $e^{\frac{e^{2 x}}{2}} e^{C}$ .

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

Step 5.2

Reorder $e^{\frac{e^{2 x}}{2}}$ and $e^{C}$ .

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

Step 5.3

Combine constants with the plus or minus.

$y = C | x | e^{\frac{e^{2 x}}{2}}$