Calculus Examples

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

Step 1

Separate the variables.

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Step 1.1

Subtract $\frac{1}{x}$ from both sides of the equation.

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

Step 1.2

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

Apply the constant rule.

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

Step 2.3

Integrate the right side.

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Step 2.3.1

Split the single integral into multiple integrals.

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

Step 2.3.2

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

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

Step 2.3.3

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

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

Step 2.3.4

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

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

Step 2.3.5

Simplify.

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

Step 2.4

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

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