Calculus Examples

$\frac{d s}{d t} = \ln (t) + 4 t$

Step 1

Rewrite the equation.

$d s = (\ln (t) + 4 t) d t$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int d s = \int \ln (t) + 4 t d t$

Step 2.2

Apply the constant rule.

$s + C_{1} = \int \ln (t) + 4 t d t$

Step 2.3

Integrate the right side.

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Step 2.3.1

Split the single integral into multiple integrals.

$s + C_{1} = \int \ln (t) d t + \int 4 t d t$

Step 2.3.2

Integrate by parts using the formula $\int u d v = u v - \int v d u$ , where $u = \ln (t)$ and $d v = 1$ .

$s + C_{1} = \ln (t) t - \int t \frac{1}{t} d t + \int 4 t d t$

Step 2.3.3

Simplify.

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Step 2.3.3.1

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

$s + C_{1} = \ln (t) t - \int \frac{t}{t} d t + \int 4 t d t$

Step 2.3.3.2

Cancel the common factor of $t$ .

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Step 2.3.3.2.1

Cancel the common factor.

$s + C_{1} = \ln (t) t - \int \frac{t}{t} d t + \int 4 t d t$

Step 2.3.3.2.2

Rewrite the expression.

$s + C_{1} = \ln (t) t - \int d t + \int 4 t d t$

Step 2.3.4

Apply the constant rule.

$s + C_{1} = \ln (t) t - (t + C_{2}) + \int 4 t d t$

Step 2.3.5

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

$s + C_{1} = \ln (t) t - (t + C_{2}) + 4 \int t d t$

Step 2.3.6

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

$s + C_{1} = \ln (t) t - (t + C_{2}) + 4 (\frac{1}{2} t^{2} + C_{3})$

Step 2.3.7

Simplify.

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Step 2.3.7.1

Simplify.

$s + C_{1} = \ln (t) t - t + 4 (\frac{1}{2}) t^{2} + C_{4}$

Step 2.3.7.2

Simplify.

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Step 2.3.7.2.1

Combine $4$ and $\frac{1}{2}$ .

$s + C_{1} = \ln (t) t - t + \frac{4}{2} t^{2} + C_{4}$

Step 2.3.7.2.2

Cancel the common factor of $4$ and $2$ .

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Step 2.3.7.2.2.1

Factor $2$ out of $4$ .

$s + C_{1} = \ln (t) t - t + \frac{2 \cdot 2}{2} t^{2} + C_{4}$

Step 2.3.7.2.2.2

Cancel the common factors.

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Step 2.3.7.2.2.2.1

Factor $2$ out of $2$ .

$s + C_{1} = \ln (t) t - t + \frac{2 \cdot 2}{2 (1)} t^{2} + C_{4}$

Step 2.3.7.2.2.2.2

Cancel the common factor.

$s + C_{1} = \ln (t) t - t + \frac{2 \cdot 2}{2 \cdot 1} t^{2} + C_{4}$

Step 2.3.7.2.2.2.3

Rewrite the expression.

$s + C_{1} = \ln (t) t - t + \frac{2}{1} t^{2} + C_{4}$

Step 2.3.7.2.2.2.4

Divide $2$ by $1$ .

$s + C_{1} = \ln (t) t - t + 2 t^{2} + C_{4}$

Step 2.3.8

Reorder terms.

$s + C_{1} = 2 t^{2} + t \ln (t) - t + C_{4}$

Step 2.4

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

$s = 2 t^{2} + t \ln (t) - t + K$