Calculus Examples

$\frac{d y}{d x} = x^{3} \ln (x)$

Step 1

Rewrite the equation.

$d y = x^{3} \ln (x) d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int d y = \int x^{3} \ln (x) d x$

Step 2.2

Apply the constant rule.

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

Step 2.3

Integrate the right side.

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Step 2.3.1

Integrate by parts using the formula $\int u d v = u v - \int v d u$ , where $u = \ln (x)$ and $d v = x^{3}$ .

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

Step 2.3.2

Simplify.

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Step 2.3.2.1

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

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

Step 2.3.2.2

Combine $\ln (x)$ and $\frac{x^{4}}{4}$ .

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

Step 2.3.3

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

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

Step 2.3.4

Simplify.

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Step 2.3.4.1

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

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

Step 2.3.4.2

Cancel the common factor of $x^{4}$ and $x$ .

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Step 2.3.4.2.1

Factor $x$ out of $x^{4}$ .

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

Step 2.3.4.2.2

Cancel the common factors.

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Step 2.3.4.2.2.1

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

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

Step 2.3.4.2.2.2

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

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

Step 2.3.4.2.2.3

Cancel the common factor.

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

Step 2.3.4.2.2.4

Rewrite the expression.

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

Step 2.3.4.2.2.5

Divide $x^{3}$ by $1$ .

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

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

Step 2.3.5

By the Power Rule, the integral of $x^{3}$ with respect to $x$ is $\frac{1}{4} x^{4}$ .

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

Step 2.3.6

Simplify the answer.

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Step 2.3.6.1

Rewrite $\frac{\ln (x) x^{4}}{4} - \frac{1}{4} (\frac{1}{4} x^{4} + C_{2})$ as $\frac{1}{4} \ln (x) x^{4} - \frac{1}{4} \cdot \frac{1}{4} x^{4} + C_{2}$ .

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

Step 2.3.6.2

Simplify.

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Step 2.3.6.2.1

Combine $\frac{1}{4}$ and $\ln (x)$ .

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

Step 2.3.6.2.2

Combine $\frac{\ln (x)}{4}$ and $x^{4}$ .

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

Step 2.3.6.2.3

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

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

Step 2.3.6.2.4

Multiply $4$ by $4$ .

$y + C_{1} = \frac{\ln (x) x^{4}}{4} - \frac{1}{16} x^{4} + C_{2}$

$y + C_{1} = \frac{1}{4} \ln (x) x^{4} - \frac{1}{16} x^{4} + C_{2}$

Step 2.3.7

Reorder terms.

$y + C_{1} = \frac{1}{4} x^{4} \ln (x) - \frac{1}{16} x^{4} + C_{2}$

Step 2.4

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

$y = \frac{1}{4} x^{4} \ln (x) - \frac{1}{16} x^{4} + K$