Calculus Examples

$\frac{d u}{d t} = 8 + 8 u + t + t u$

Step 1

Separate the variables.

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Step 1.1

Factor.

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Step 1.1.1

Factor out the greatest common factor from each group.

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Step 1.1.1.1

Group the first two terms and the last two terms.

$\frac{d u}{d t} = (8 + 8 u) + t + t u$

Step 1.1.1.2

Factor out the greatest common factor (GCF) from each group.

$\frac{d u}{d t} = 8 (1 + u) + t \cdot (1 + u)$

Step 1.1.2

Factor the polynomial by factoring out the greatest common factor, $1 + u$ .

$\frac{d u}{d t} = (1 + u) (8 + t)$

Step 1.2

Multiply both sides by $\frac{1}{1 + u}$ .

$\frac{1}{1 + u} \frac{d u}{d t} = \frac{1}{1 + u} ((1 + u) (8 + t))$

Step 1.3

Cancel the common factor of $1 + u$ .

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Step 1.3.1

Cancel the common factor.

$\frac{1}{1 + u} \frac{d u}{d t} = \frac{1}{1 + u} ((1 + u) (8 + t))$

Step 1.3.2

Rewrite the expression.

$\frac{1}{1 + u} \frac{d u}{d t} = 8 + t$

Step 1.4

Rewrite the equation.

$\frac{1}{1 + u} d u = (8 + t) d t$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int \frac{1}{1 + u} d u = \int 8 + t d t$

Step 2.2

Integrate the left side.

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Step 2.2.1

Let $u_{1} = 1 + u$ . Then $d u_{1} = d u$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

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Step 2.2.1.1

Let $u_{1} = 1 + u$ . Find $\frac{d u_{1}}{d u}$ .

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Step 2.2.1.1.1

Differentiate $1 + u$ .

$\frac{d}{d u} [1 + u]$

Step 2.2.1.1.2

By the Sum Rule, the derivative of $1 + u$ with respect to $u$ is $\frac{d}{d u} [1] + \frac{d}{d u} [u]$ .

$\frac{d}{d u} [1] + \frac{d}{d u} [u]$

Step 2.2.1.1.3

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

$0 + \frac{d}{d u} [u]$

Step 2.2.1.1.4

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

$0 + 1$

Step 2.2.1.1.5

Add $0$ and $1$ .

$1$

Step 2.2.1.2

Rewrite the problem using $u_{1}$ and $d u_{1}$ .

$\int \frac{1}{u_{1}} d u_{1} = \int 8 + t d t$

Step 2.2.2

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

$\ln (| u_{1} |) + C_{1} = \int 8 + t d t$

Step 2.2.3

Replace all occurrences of $u_{1}$ with $1 + u$ .

$\ln (| 1 + u |) + C_{1} = \int 8 + t d t$

Step 2.3

Integrate the right side.

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Step 2.3.1

Split the single integral into multiple integrals.

$\ln (| 1 + u |) + C_{1} = \int 8 d t + \int t d t$

Step 2.3.2

Apply the constant rule.

$\ln (| 1 + u |) + C_{1} = 8 t + C_{2} + \int t d t$

Step 2.3.3

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

$\ln (| 1 + u |) + C_{1} = 8 t + C_{2} + \frac{1}{2} t^{2} + C_{3}$

Step 2.3.4

Simplify.

$\ln (| 1 + u |) + C_{1} = 8 t + \frac{1}{2} t^{2} + C_{4}$

Step 2.3.5

Reorder terms.

$\ln (| 1 + u |) + C_{1} = \frac{1}{2} t^{2} + 8 t + C_{4}$

Step 2.4

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

$\ln (| 1 + u |) = \frac{1}{2} t^{2} + 8 t + C$

Step 3

Solve for $u$ .

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Step 3.1

To solve for $u$ , rewrite the equation using properties of logarithms.

$e^{\ln (| 1 + u |)} = e^{\frac{1}{2} t^{2} + 8 t + C}$

Step 3.2

Rewrite $\ln (| 1 + u |) = \frac{1}{2} t^{2} + 8 t + C$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b \neq 1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$e^{\frac{1}{2} t^{2} + 8 t + C} = | 1 + u |$

Step 3.3

Solve for $u$ .

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Step 3.3.1

Rewrite the equation as $| 1 + u | = e^{\frac{1}{2} t^{2} + 8 t + C}$ .

$| 1 + u | = e^{\frac{1}{2} t^{2} + 8 t + C}$

Step 3.3.2

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

$| 1 + u | = e^{\frac{t^{2}}{2} + 8 t + C}$

Step 3.3.3

Remove the absolute value term. This creates a $\pm$ on the right side of the equation because $| x | = \pm x$ .

$1 + u = \pm e^{\frac{t^{2}}{2} + 8 t + C}$

Step 3.3.4

Subtract $1$ from both sides of the equation.

$u = \pm e^{\frac{t^{2}}{2} + 8 t + C} - 1$

Step 4

Group the constant terms together.

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Step 4.1

Rewrite $e^{\frac{t^{2}}{2} + 8 t + C}$ as $e^{\frac{t^{2}}{2} + 8 t} e^{C}$ .

$u = \pm e^{\frac{t^{2}}{2} + 8 t} e^{C} - 1$

Step 4.2

Reorder $e^{\frac{t^{2}}{2} + 8 t}$ and $e^{C}$ .

$u = \pm e^{C} e^{\frac{t^{2}}{2} + 8 t} - 1$

Step 4.3

Combine constants with the plus or minus.

$u = C e^{\frac{t^{2}}{2} + 8 t} - 1$