Calculus Examples

f (x) = \frac{2}{x + 1}

$f (x) = \frac{2}{x + 1}$

Step 1

The function

F (x)

$F (x)$ can be found by finding the indefinite integral of the derivative

f (x)

$f (x)$ .

F (x) = \int f (x) d x

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

Step 2

Set up the integral to solve.

F (x) = \int \frac{2}{x + 1} d x

$F (x) = \int \frac{2}{x + 1} d x$

Step 3

Since

2

$2$ is constant with respect to

x

$x$ , move

2

$2$ out of the integral.

2 \int \frac{1}{x + 1} d x

$2 \int \frac{1}{x + 1} d x$

Step 4

Let

u = x + 1

$u = x + 1$ . Then

d u = d x

$d u = d x$ . Rewrite using

u

$u$ and

d

$d$

u

$u$ .

Tap for more steps...

Step 4.1

Let

u = x + 1

$u = x + 1$ . Find

\frac{d u}{d x}

$\frac{d u}{d x}$ .

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Step 4.1.1

Differentiate

x + 1

$x + 1$ .

\frac{d}{d x} [x + 1]

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

Step 4.1.2

By the Sum Rule, the derivative of

x + 1

$x + 1$ with respect to

x

$x$ is

\frac{d}{d x} [x] + \frac{d}{d x} [1]

$\frac{d}{d x} [x] + \frac{d}{d x} [1]$ .

\frac{d}{d x} [x] + \frac{d}{d x} [1]

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

Step 4.1.3

Differentiate using the Power Rule which states that

\frac{d}{d x} [x^{n}]

$\frac{d}{d x} [x^{n}]$ is

n x^{n - 1}

$n x^{n - 1}$ where

n = 1

$n = 1$ .

1 + \frac{d}{d x} [1]

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

Step 4.1.4

Since

1

$1$ is constant with respect to

x

$x$ , the derivative of

1

$1$ with respect to

x

$x$ is

0

$0$ .

1 + 0

$1 + 0$

Step 4.1.5

Add

1

$1$ and

0

$0$ .

1

$1$

1

$1$

Step 4.2

Rewrite the problem using

u

$u$ and

d u

$d u$ .

2 \int \frac{1}{u} d u

$2 \int \frac{1}{u} d u$

2 \int \frac{1}{u} d u

$2 \int \frac{1}{u} d u$

Step 5

The integral of

\frac{1}{u}

$\frac{1}{u}$ with respect to

u

$u$ is

ln (| u |)

$\ln (| u |)$ .

2 (ln (| u |) + C)

$2 (\ln (| u |) + C)$

Step 6

Simplify.

2 ln (| u |) + C

$2 \ln (| u |) + C$

Step 7

Replace all occurrences of

u

$u$ with

x + 1

$x + 1$ .

2 ln (| x + 1 |) + C

$2 \ln (| x + 1 |) + C$

Step 8

The answer is the antiderivative of the function

f (x) = \frac{2}{x + 1}

$f (x) = \frac{2}{x + 1}$ .

F (x) =

$F (x) =$

2 ln (| x + 1 |) + C

$2 \ln (| x + 1 |) + C$