Calculus Examples

$e^{2 x} + \frac{1}{x + 1}$

Step 1

Write $e^{2 x} + \frac{1}{x + 1}$ as a function.

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

Step 2

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

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

Step 3

Set up the integral to solve.

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

Step 4

Split the single integral into multiple integrals.

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

Step 5

Let $u_{1} = 2 x$ . Then $d u_{1} = 2 d x$ , so $\frac{1}{2} d u_{1} = d x$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

Tap for more steps...

Step 5.1

Let $u_{1} = 2 x$ . Find $\frac{d u_{1}}{d x}$ .

Tap for more steps...

Step 5.1.1

Differentiate $2 x$ .

$\frac{d}{d x} [2 x]$

Step 5.1.2

Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

$2 \frac{d}{d x} [x]$

Step 5.1.3

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

$2 \cdot 1$

Step 5.1.4

Multiply $2$ by $1$ .

$2$

Step 5.2

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

$\int e^{u_{1}} \frac{1}{2} d u_{1} + \int \frac{1}{x + 1} d x$

Step 6

Combine $e^{u_{1}}$ and $\frac{1}{2}$ .

$\int \frac{e^{u_{1}}}{2} d u_{1} + \int \frac{1}{x + 1} d x$

Step 7

Since $\frac{1}{2}$ is constant with respect to $u_{1}$ , move $\frac{1}{2}$ out of the integral.

$\frac{1}{2} \int e^{u_{1}} d u_{1} + \int \frac{1}{x + 1} d x$

Step 8

The integral of $e^{u_{1}}$ with respect to $u_{1}$ is $e^{u_{1}}$ .

$\frac{1}{2} (e^{u_{1}} + C) + \int \frac{1}{x + 1} d x$

Step 9

Let $u_{2} = x + 1$ . Then $d u_{2} = d x$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

Tap for more steps...

Step 9.1

Let $u_{2} = x + 1$ . Find $\frac{d u_{2}}{d x}$ .

Tap for more steps...

Step 9.1.1

Differentiate $x + 1$ .

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

Step 9.1.2

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

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

Step 9.1.3

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

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

Step 9.1.4

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

$1 + 0$

Step 9.1.5

Add $1$ and $0$ .

$1$

Step 9.2

Rewrite the problem using $u_{2}$ and $d u_{2}$ .

$\frac{1}{2} (e^{u_{1}} + C) + \int \frac{1}{u_{2}} d u_{2}$

Step 10

The integral of $\frac{1}{u_{2}}$ with respect to $u_{2}$ is $\ln (| u_{2} |)$ .

$\frac{1}{2} (e^{u_{1}} + C) + \ln (| u_{2} |) + C$

Step 11

Simplify.

$\frac{1}{2} e^{u_{1}} + \ln (| u_{2} |) + C$

Step 12

Substitute back in for each integration substitution variable.

Tap for more steps...

Step 12.1

Replace all occurrences of $u_{1}$ with $2 x$ .

$\frac{1}{2} e^{2 x} + \ln (| u_{2} |) + C$

Step 12.2

Replace all occurrences of $u_{2}$ with $x + 1$ .

$\frac{1}{2} e^{2 x} + \ln (| x + 1 |) + C$

Step 13

The answer is the antiderivative of the function $f (x) = e^{2 x} + \frac{1}{x + 1}$ .

$F (x) =$ $\frac{1}{2} e^{2 x} + \ln (| x + 1 |) + C$