Calculus Examples

$(6 x - 5) \frac{d x}{d y} = \frac{- 3 x^{2} + 5 x + 1}{y}$

Step 1

Separate the variables.

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Step 1.1

Multiply both sides by $\frac{1}{- 3 x^{2} + 5 x + 1}$ .

$\frac{1}{- 3 x^{2} + 5 x + 1} ((6 x - 5) \frac{d x}{d y}) = \frac{1}{- 3 x^{2} + 5 x + 1} \cdot \frac{- 3 x^{2} + 5 x + 1}{y}$

Step 1.2

Simplify.

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Step 1.2.1

Multiply $\frac{1}{- 3 x^{2} + 5 x + 1}$ by $\frac{- 3 x^{2} + 5 x + 1}{y}$ .

$\frac{1}{- 3 x^{2} + 5 x + 1} ((6 x - 5) \frac{d x}{d y}) = \frac{- 3 x^{2} + 5 x + 1}{(- 3 x^{2} + 5 x + 1) y}$

Step 1.2.2

Cancel the common factor of $- 3 x^{2} + 5 x + 1$ .

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Step 1.2.2.1

Cancel the common factor.

$\frac{1}{- 3 x^{2} + 5 x + 1} ((6 x - 5) \frac{d x}{d y}) = \frac{- 3 x^{2} + 5 x + 1}{(- 3 x^{2} + 5 x + 1) y}$

Step 1.2.2.2

Rewrite the expression.

$\frac{1}{- 3 x^{2} + 5 x + 1} ((6 x - 5) \frac{d x}{d y}) = \frac{1}{y}$

Step 1.3

Remove unnecessary parentheses.

$\frac{1}{- 3 x^{2} + 5 x + 1} (6 x - 5) \frac{d x}{d y} = \frac{1}{y}$

Step 1.4

Rewrite the equation.

$\frac{1}{- 3 x^{2} + 5 x + 1} (6 x - 5) d x = \frac{1}{y} d y$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int \frac{1}{- 3 x^{2} + 5 x + 1} (6 x - 5) d x = \int \frac{1}{y} d y$

Step 2.2

Integrate the left side.

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Step 2.2.1

Let $u = - 3 x^{2} + 5 x + 1$ . Then $d u = (- 6 x + 5) d x$ , so $- d u = (6 x - 5) d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 2.2.1.1

Let $u = - 3 x^{2} + 5 x + 1$ . Find $\frac{d u}{d x}$ .

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Step 2.2.1.1.1

Differentiate $- 3 x^{2} + 5 x + 1$ .

$\frac{d}{d x} [- 3 x^{2} + 5 x + 1]$

Step 2.2.1.1.2

By the Sum Rule, the derivative of $- 3 x^{2} + 5 x + 1$ with respect to $x$ is $\frac{d}{d x} [- 3 x^{2}] + \frac{d}{d x} [5 x] + \frac{d}{d x} [1]$ .

$\frac{d}{d x} [- 3 x^{2}] + \frac{d}{d x} [5 x] + \frac{d}{d x} [1]$

Step 2.2.1.1.3

Evaluate $\frac{d}{d x} [- 3 x^{2}]$ .

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Step 2.2.1.1.3.1

Since $- 3$ is constant with respect to $x$ , the derivative of $- 3 x^{2}$ with respect to $x$ is $- 3 \frac{d}{d x} [x^{2}]$ .

$- 3 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [5 x] + \frac{d}{d x} [1]$

Step 2.2.1.1.3.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$- 3 (2 x) + \frac{d}{d x} [5 x] + \frac{d}{d x} [1]$

Step 2.2.1.1.3.3

Multiply $2$ by $- 3$ .

$- 6 x + \frac{d}{d x} [5 x] + \frac{d}{d x} [1]$

Step 2.2.1.1.4

Evaluate $\frac{d}{d x} [5 x]$ .

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Step 2.2.1.1.4.1

Since $5$ is constant with respect to $x$ , the derivative of $5 x$ with respect to $x$ is $5 \frac{d}{d x} [x]$ .

$- 6 x + 5 \frac{d}{d x} [x] + \frac{d}{d x} [1]$

Step 2.2.1.1.4.2

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

$- 6 x + 5 \cdot 1 + \frac{d}{d x} [1]$

Step 2.2.1.1.4.3

Multiply $5$ by $1$ .

$- 6 x + 5 + \frac{d}{d x} [1]$

Step 2.2.1.1.5

Differentiate using the Constant Rule.

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Step 2.2.1.1.5.1

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$- 6 x + 5 + 0$

Step 2.2.1.1.5.2

Add $- 6 x + 5$ and $0$ .

$- 6 x + 5$

Step 2.2.1.2

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

$\int \frac{- 1}{u} d u = \int \frac{1}{y} d y$

Step 2.2.2

Split the fraction into multiple fractions.

$\int - \frac{1}{u} d u = \int \frac{1}{y} d y$

Step 2.2.3

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

$- \int \frac{1}{u} d u = \int \frac{1}{y} d y$

Step 2.2.4

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

$- (\ln (| u |) + C_{1}) = \int \frac{1}{y} d y$

Step 2.2.5

Simplify.

$- \ln (| u |) + C_{1} = \int \frac{1}{y} d y$

Step 2.2.6

Replace all occurrences of $u$ with $- 3 x^{2} + 5 x + 1$ .

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

Step 2.3

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

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

Step 2.4

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

$- \ln (| - 3 x^{2} + 5 x + 1 |) = \ln (| y |) + C$