Calculus Examples

$x \frac{d y}{d x} = 5 x^{3} + 4$

Step 1

Separate the variables.

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Step 1.1

Divide each term in $x \frac{d y}{d x} = 5 x^{3} + 4$ by $x$ and simplify.

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Step 1.1.1

Divide each term in $x \frac{d y}{d x} = 5 x^{3} + 4$ by $x$ .

$\frac{x \frac{d y}{d x}}{x} = \frac{5 x^{3}}{x} + \frac{4}{x}$

Step 1.1.2

Simplify the left side.

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Step 1.1.2.1

Cancel the common factor of $x$ .

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Step 1.1.2.1.1

Cancel the common factor.

$\frac{x \frac{d y}{d x}}{x} = \frac{5 x^{3}}{x} + \frac{4}{x}$

Step 1.1.2.1.2

Divide $\frac{d y}{d x}$ by $1$ .

$\frac{d y}{d x} = \frac{5 x^{3}}{x} + \frac{4}{x}$

Step 1.1.3

Simplify the right side.

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Step 1.1.3.1

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

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Step 1.1.3.1.1

Factor $x$ out of $5 x^{3}$ .

$\frac{d y}{d x} = \frac{x (5 x^{2})}{x} + \frac{4}{x}$

Step 1.1.3.1.2

Cancel the common factors.

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Step 1.1.3.1.2.1

Raise $x$ to the power of $1$ .

$\frac{d y}{d x} = \frac{x (5 x^{2})}{x^{1}} + \frac{4}{x}$

Step 1.1.3.1.2.2

Factor $x$ out of $x^{1}$ .

$\frac{d y}{d x} = \frac{x (5 x^{2})}{x \cdot 1} + \frac{4}{x}$

Step 1.1.3.1.2.3

Cancel the common factor.

$\frac{d y}{d x} = \frac{x (5 x^{2})}{x \cdot 1} + \frac{4}{x}$

Step 1.1.3.1.2.4

Rewrite the expression.

$\frac{d y}{d x} = \frac{5 x^{2}}{1} + \frac{4}{x}$

Step 1.1.3.1.2.5

Divide $5 x^{2}$ by $1$ .

$\frac{d y}{d x} = 5 x^{2} + \frac{4}{x}$

Step 1.2

Rewrite the equation.

$d y = (5 x^{2} + \frac{4}{x}) d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int d y = \int 5 x^{2} + \frac{4}{x} d x$

Step 2.2

Apply the constant rule.

$y + C_{1} = \int 5 x^{2} + \frac{4}{x} d x$

Step 2.3

Integrate the right side.

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Step 2.3.1

Split the single integral into multiple integrals.

$y + C_{1} = \int 5 x^{2} d x + \int \frac{4}{x} d x$

Step 2.3.2

Since $5$ is constant with respect to $x$ , move $5$ out of the integral.

$y + C_{1} = 5 \int x^{2} d x + \int \frac{4}{x} d x$

Step 2.3.3

By the Power Rule, the integral of $x^{2}$ with respect to $x$ is $\frac{1}{3} x^{3}$ .

$y + C_{1} = 5 (\frac{1}{3} x^{3} + C_{2}) + \int \frac{4}{x} d x$

Step 2.3.4

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

$y + C_{1} = 5 (\frac{1}{3} x^{3} + C_{2}) + 4 \int \frac{1}{x} d x$

Step 2.3.5

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

$y + C_{1} = 5 (\frac{1}{3} x^{3} + C_{2}) + 4 (\ln (| x |) + C_{3})$

Step 2.3.6

Simplify.

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Step 2.3.6.1

Simplify.

$y + C_{1} = 5 (\frac{1}{3}) x^{3} + 4 \ln (| x |) + C_{4}$

Step 2.3.6.2

Combine $5$ and $\frac{1}{3}$ .

$y + C_{1} = \frac{5}{3} x^{3} + 4 \ln (| x |) + C_{4}$

Step 2.4

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

$y = \frac{5}{3} x^{3} + 4 \ln (| x |) + C$