Calculus Examples

$\frac{d L}{d t} = {kL}^{2} \ln (t)$

Step 1

Rewrite the equation.

$d L = {kL}^{2} \ln (t) d t$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int d L = \int {kL}^{2} \ln (t) d t$

Step 2.2

Apply the constant rule.

$L + C_{1} = \int {kL}^{2} \ln (t) d t$

Step 2.3

Integrate the right side.

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Step 2.3.1

Since ${kL}^{2}$ is constant with respect to $t$ , move ${kL}^{2}$ out of the integral.

$L + C_{1} = {kL}^{2} \int \ln (t) d t$

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 $d v = 1$ .

$L + C_{1} = {kL}^{2} (\ln (t) t - \int t \frac{1}{t} d t)$

Step 2.3.3

Simplify.

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Step 2.3.3.1

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

$L + C_{1} = {kL}^{2} (\ln (t) t - \int \frac{t}{t} d t)$

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.

$L + C_{1} = {kL}^{2} (\ln (t) t - \int \frac{t}{t} d t)$

Step 2.3.3.2.2

Rewrite the expression.

$L + C_{1} = {kL}^{2} (\ln (t) t - \int d t)$

Step 2.3.4

Apply the constant rule.

$L + C_{1} = {kL}^{2} (\ln (t) t - (t + C_{2}))$

Step 2.3.5

Simplify.

$L + C_{1} = {kL}^{2} (\ln (t) t - t) + C_{2}$

Step 2.4

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

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