Calculus Examples

$\frac{d x}{d t} = \frac{t - 1}{x^{2} - 4 x + 4}$

Step 1

Separate the variables.

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Step 1.1

Multiply both sides by $x^{2} - 4 x + 4$ .

$(x^{2} - 4 x + 4) \frac{d x}{d t} = (x^{2} - 4 x + 4) \frac{t - 1}{x^{2} - 4 x + 4}$

Step 1.2

Simplify.

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Step 1.2.1

Factor using the perfect square rule.

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Step 1.2.1.1

Rewrite $4$ as $2^{2}$ .

$(x^{2} - 4 x + 4) \frac{d x}{d t} = (x^{2} - 4 x + 4) \frac{t - 1}{x^{2} - 4 x + 2^{2}}$

Step 1.2.1.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$4 x = 2 \cdot x \cdot 2$

Step 1.2.1.3

Rewrite the polynomial.

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

Step 1.2.1.4

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 2$ .

$(x^{2} - 4 x + 4) \frac{d x}{d t} = (x^{2} - 4 x + 4) \frac{t - 1}{{(x - 2)}^{2}}$

Step 1.2.2

Multiply $x^{2} - 4 x + 4$ by $\frac{t - 1}{{(x - 2)}^{2}}$ .

$(x^{2} - 4 x + 4) \frac{d x}{d t} = \frac{(x^{2} - 4 x + 4) (t - 1)}{{(x - 2)}^{2}}$

Step 1.2.3

Factor using the perfect square rule.

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Step 1.2.3.1

Rewrite $4$ as $2^{2}$ .

$(x^{2} - 4 x + 4) \frac{d x}{d t} = \frac{(x^{2} - 4 x + 2^{2}) (t - 1)}{{(x - 2)}^{2}}$

Step 1.2.3.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$4 x = 2 \cdot x \cdot 2$

Step 1.2.3.3

Rewrite the polynomial.

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

Step 1.2.3.4

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 2$ .

$(x^{2} - 4 x + 4) \frac{d x}{d t} = \frac{{(x - 2)}^{2} (t - 1)}{{(x - 2)}^{2}}$

Step 1.2.4

Cancel the common factor of ${(x - 2)}^{2}$ .

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Step 1.2.4.1

Cancel the common factor.

$(x^{2} - 4 x + 4) \frac{d x}{d t} = \frac{{(x - 2)}^{2} (t - 1)}{{(x - 2)}^{2}}$

Step 1.2.4.2

Divide $t - 1$ by $1$ .

$(x^{2} - 4 x + 4) \frac{d x}{d t} = t - 1$

Step 1.3

Rewrite the equation.

$(x^{2} - 4 x + 4) d x = (t - 1) d t$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int x^{2} - 4 x + 4 d x = \int t - 1 d t$

Step 2.2

Integrate the left side.

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Step 2.2.1

Split the single integral into multiple integrals.

$\int x^{2} d x + \int - 4 x d x + \int 4 d x = \int t - 1 d t$

Step 2.2.2

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

$\frac{1}{3} x^{3} + C_{1} + \int - 4 x d x + \int 4 d x = \int t - 1 d t$

Step 2.2.3

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

$\frac{1}{3} x^{3} + C_{1} - 4 \int x d x + \int 4 d x = \int t - 1 d t$

Step 2.2.4

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

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

Step 2.2.5

Apply the constant rule.

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

Step 2.2.6

Simplify.

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Step 2.2.6.1

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

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

Step 2.2.6.2

Simplify.

$\frac{x^{3}}{3} - 2 x^{2} + 4 x + C_{4} = \int t - 1 d t$

Step 2.2.6.3

Reorder terms.

$\frac{1}{3} x^{3} - 2 x^{2} + 4 x + C_{4} = \int t - 1 d t$

Step 2.2.7

Reorder terms.

$- 2 x^{2} + \frac{1}{3} x^{3} + 4 x + C_{4} = \int t - 1 d t$

Step 2.3

Integrate the right side.

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Step 2.3.1

Split the single integral into multiple integrals.

$- 2 x^{2} + \frac{1}{3} x^{3} + 4 x + C_{4} = \int t d t + \int - 1 d t$

Step 2.3.2

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

$- 2 x^{2} + \frac{1}{3} x^{3} + 4 x + C_{4} = \frac{1}{2} t^{2} + C_{5} + \int - 1 d t$

Step 2.3.3

Apply the constant rule.

$- 2 x^{2} + \frac{1}{3} x^{3} + 4 x + C_{4} = \frac{1}{2} t^{2} + C_{5} - t + C_{6}$

Step 2.3.4

Simplify.

$- 2 x^{2} + \frac{1}{3} x^{3} + 4 x + C_{4} = \frac{1}{2} t^{2} - t + C_{7}$

Step 2.4

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

$- 2 x^{2} + \frac{1}{3} x^{3} + 4 x = \frac{1}{2} t^{2} - t + K$