Calculus Examples

$(2 x + 5) d x - d y = 0$

Step 1

Subtract $(2 x + 5) d x$ from both sides of the equation.

$- d y = - (2 x + 5) d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int - 1 d y = \int - (2 x + 5) d x$

Step 2.2

Apply the constant rule.

$- y + C_{1} = \int - (2 x + 5) d x$

Step 2.3

Integrate the right side.

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Step 2.3.1

Multiply $- (2 x + 5)$ .

$- y + C_{1} = \int - (2 x) - 1 \cdot 5 d x$

Step 2.3.2

Simplify.

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Step 2.3.2.1

Multiply $2$ by $- 1$ .

$- y + C_{1} = \int - 2 x - 1 \cdot 5 d x$

Step 2.3.2.2

Multiply $- 1$ by $5$ .

$- y + C_{1} = \int - 2 x - 5 d x$

Step 2.3.3

Split the single integral into multiple integrals.

$- y + C_{1} = \int - 2 x d x + \int - 5 d x$

Step 2.3.4

Since $- 2$ is constant with respect to $x$ , move $- 2$ out of the integral.

$- y + C_{1} = - 2 \int x d x + \int - 5 d x$

Step 2.3.5

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

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

Step 2.3.6

Apply the constant rule.

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

Step 2.3.7

Simplify.

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Step 2.3.7.1

Combine $\frac{1}{2}$ and $x^{2}$ .

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

Step 2.3.7.2

Simplify.

$- y + C_{1} = - x^{2} - 5 x + C_{4}$

Step 2.4

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

$- y = - x^{2} - 5 x + K$

Step 3

Divide each term in $- y = - x^{2} - 5 x + K$ by $- 1$ and simplify.

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Step 3.1

Divide each term in $- y = - x^{2} - 5 x + K$ by $- 1$ .

$\frac{- y}{- 1} = \frac{- x^{2}}{- 1} + \frac{- 5 x}{- 1} + \frac{K}{- 1}$

Step 3.2

Simplify the left side.

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Step 3.2.1

Dividing two negative values results in a positive value.

$\frac{y}{1} = \frac{- x^{2}}{- 1} + \frac{- 5 x}{- 1} + \frac{K}{- 1}$

Step 3.2.2

Divide $y$ by $1$ .

$y = \frac{- x^{2}}{- 1} + \frac{- 5 x}{- 1} + \frac{K}{- 1}$

Step 3.3

Simplify the right side.

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Step 3.3.1

Simplify each term.

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Step 3.3.1.1

Dividing two negative values results in a positive value.

$y = \frac{x^{2}}{1} + \frac{- 5 x}{- 1} + \frac{K}{- 1}$

Step 3.3.1.2

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

$y = x^{2} + \frac{- 5 x}{- 1} + \frac{K}{- 1}$

Step 3.3.1.3

Move the negative one from the denominator of $\frac{- 5 x}{- 1}$ .

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

Step 3.3.1.4

Rewrite $- 1 \cdot (- 5 x)$ as $- (- 5 x)$ .

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

Step 3.3.1.5

Multiply $- 5$ by $- 1$ .

$y = x^{2} + 5 x + \frac{K}{- 1}$

Step 3.3.1.6

Move the negative one from the denominator of $\frac{K}{- 1}$ .

$y = x^{2} + 5 x - 1 \cdot K$

Step 3.3.1.7

Rewrite $- 1 \cdot K$ as $- K$ .

$y = x^{2} + 5 x - K$

Step 4

Simplify the constant of integration.

$y = x^{2} + 5 x + K$