Calculus Examples

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

Step 1

Separate the variables.

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Step 1.1

Solve for $\frac{d y}{d x}$ .

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Step 1.1.1

Simplify.

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Step 1.1.1.1

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

$f (x^{1} x^{3}) \frac{d y}{d x} = \ln (x)$

Step 1.1.1.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f x^{1 + 3} \frac{d y}{d x} = \ln (x)$

Step 1.1.1.3

Add $1$ and $3$ .

$f x^{4} \frac{d y}{d x} = \ln (x)$

Step 1.1.2

Divide each term in $f x^{4} \frac{d y}{d x} = \ln (x)$ by $f x^{4}$ and simplify.

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Step 1.1.2.1

Divide each term in $f x^{4} \frac{d y}{d x} = \ln (x)$ by $f x^{4}$ .

$\frac{f x^{4} \frac{d y}{d x}}{f x^{4}} = \frac{\ln (x)}{f x^{4}}$

Step 1.1.2.2

Simplify the left side.

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Step 1.1.2.2.1

Cancel the common factor of $f$ .

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Step 1.1.2.2.1.1

Cancel the common factor.

$\frac{f x^{4} \frac{d y}{d x}}{f x^{4}} = \frac{\ln (x)}{f x^{4}}$

Step 1.1.2.2.1.2

Rewrite the expression.

$\frac{x^{4} \frac{d y}{d x}}{x^{4}} = \frac{\ln (x)}{f x^{4}}$

Step 1.1.2.2.2

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

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Step 1.1.2.2.2.1

Cancel the common factor.

$\frac{x^{4} \frac{d y}{d x}}{x^{4}} = \frac{\ln (x)}{f x^{4}}$

Step 1.1.2.2.2.2

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

$\frac{d y}{d x} = \frac{\ln (x)}{f x^{4}}$

Step 1.2

Rewrite the equation.

$d y = \frac{\ln (x)}{f x^{4}} d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int d y = \int \frac{\ln (x)}{f x^{4}} d x$

Step 2.2

Apply the constant rule.

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

Step 2.3

Integrate the right side.

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Step 2.3.1

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

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

Step 2.3.2

Apply basic rules of exponents.

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Step 2.3.2.1

Move $x^{4}$ out of the denominator by raising it to the $- 1$ power.

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

Step 2.3.2.2

Multiply the exponents in ${(x^{4})}^{- 1}$ .

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Step 2.3.2.2.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Step 2.3.2.2.2

Multiply $4$ by $- 1$ .

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

Step 2.3.3

Integrate by parts using the formula $\int u d v = u v - \int v d u$ , where $u = \ln (x)$ and $d v = x^{- 4}$ .

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

Step 2.3.4

Simplify.

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Step 2.3.4.1

Combine $\ln (x)$ and $\frac{1}{3 x^{3}}$ .

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

Step 2.3.4.2

Multiply $\frac{1}{x}$ by $\frac{1}{3 x^{3}}$ .

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

Step 2.3.4.3

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

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

Step 2.3.4.4

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 2.3.4.5

Add $1$ and $3$ .

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

Step 2.3.5

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

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

Step 2.3.6

Simplify.

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Step 2.3.6.1

Multiply $- 1$ by $- 1$ .

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

Step 2.3.6.2

Multiply $\int \frac{1}{3 x^{4}} d x$ by $1$ .

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

Step 2.3.7

Since $\frac{1}{3}$ is constant with respect to $x$ , move $\frac{1}{3}$ out of the integral.

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

Step 2.3.8

Apply basic rules of exponents.

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Step 2.3.8.1

Move $x^{4}$ out of the denominator by raising it to the $- 1$ power.

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

Step 2.3.8.2

Multiply the exponents in ${(x^{4})}^{- 1}$ .

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Step 2.3.8.2.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Step 2.3.8.2.2

Multiply $4$ by $- 1$ .

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

Step 2.3.9

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

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

Step 2.3.10

Simplify the answer.

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Step 2.3.10.1

Simplify.

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Step 2.3.10.1.1

Combine $x^{- 3}$ and $\frac{1}{3}$ .

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

Step 2.3.10.1.2

Move $x^{- 3}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Step 2.3.10.2

Rewrite $\frac{1}{f} (- \frac{\ln (x)}{3 x^{3}} + \frac{1}{3} (- \frac{1}{3 x^{3}} + C_{2}))$ as $\frac{1}{f} (- \frac{\ln (x)}{3 x^{3}} - \frac{1}{9 x^{3}}) + C_{2}$ .

$y + C_{1} = \frac{1}{f} (- \frac{\ln (x)}{3 x^{3}} - \frac{1}{9 x^{3}}) + C_{2}$

Step 2.3.11

Reorder terms.

$y + C_{1} = (- \frac{\ln (x)}{3 x^{3}} - \frac{1}{9 x^{3}}) \frac{1}{f} + C_{2}$

Step 2.4

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

$y = (- \frac{\ln (x)}{3 x^{3}} - \frac{1}{9 x^{3}}) \frac{1}{f} + K$