Calculus Examples

$\frac{20}{6 + 2 t}$

Step 1

Write $\frac{20}{6 + 2 t}$ as a function.

$f (t) = \frac{20}{6 + 2 t}$

Step 2

The function $F (t)$ can be found by finding the indefinite integral of the derivative $f (t)$ .

$F (t) = \int f (t) d t$

Step 3

Set up the integral to solve.

$F (t) = \int \frac{20}{6 + 2 t} d t$

Step 4

Cancel the common factor of $20$ and $6 + 2 t$ .

Tap for more steps...

Step 4.1

Factor $2$ out of $20$ .

$\int \frac{2 \cdot 10}{6 + 2 t} d t$

Step 4.2

Cancel the common factors.

Tap for more steps...

Step 4.2.1

Factor $2$ out of $6$ .

$\int \frac{2 \cdot 10}{2 (3) + 2 t} d t$

Step 4.2.2

Factor $2$ out of $2 t$ .

$\int \frac{2 \cdot 10}{2 (3) + 2 (t)} d t$

Step 4.2.3

Factor $2$ out of $2 (3) + 2 (t)$ .

$\int \frac{2 \cdot 10}{2 (3 + t)} d t$

Step 4.2.4

Cancel the common factor.

$\int \frac{2 \cdot 10}{2 (3 + t)} d t$

Step 4.2.5

Rewrite the expression.

$\int \frac{10}{3 + t} d t$

Step 5

Since $10$ is constant with respect to $t$ , move $10$ out of the integral.

$10 \int \frac{1}{3 + t} d t$

Step 6

Let $u = 3 + t$ . Then $d u = d t$ . Rewrite using $u$ and $d$ $u$ .

Tap for more steps...

Step 6.1

Let $u = 3 + t$ . Find $\frac{d u}{d t}$ .

Tap for more steps...

Step 6.1.1

Differentiate $3 + t$ .

$\frac{d}{d t} [3 + t]$

Step 6.1.2

By the Sum Rule, the derivative of $3 + t$ with respect to $t$ is $\frac{d}{d t} [3] + \frac{d}{d t} [t]$ .

$\frac{d}{d t} [3] + \frac{d}{d t} [t]$

Step 6.1.3

Since $3$ is constant with respect to $t$ , the derivative of $3$ with respect to $t$ is $0$ .

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

Step 6.1.4

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

$0 + 1$

Step 6.1.5

Add $0$ and $1$ .

$1$

Step 6.2

Rewrite the problem using $u$ and $d u$ .

$10 \int \frac{1}{u} d u$

Step 7

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

$10 (\ln (| u |) + C)$

Step 8

Simplify.

$10 \ln (| u |) + C$

Step 9

Replace all occurrences of $u$ with $3 + t$ .

$10 \ln (| 3 + t |) + C$

Step 10

The answer is the antiderivative of the function $f (t) = \frac{20}{6 + 2 t}$ .

$F (t) =$ $10 \ln (| 3 + t |) + C$