Calculus Examples

$\frac{4}{\sqrt{x}} + \frac{3}{x} - 2 e^{- x}$

Step 1

Write $\frac{4}{\sqrt{x}} + \frac{3}{x} - 2 e^{- x}$ as a function.

$f (x) = \frac{4}{\sqrt{x}} + \frac{3}{x} - 2 e^{- x}$

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.

$F (x) = \int \frac{4}{\sqrt{x}} + \frac{3}{x} - 2 e^{- x} d x$

Step 4

Split the single integral into multiple integrals.

$\int \frac{4}{\sqrt{x}} d x + \int \frac{3}{x} d x + \int - 2 e^{- x} d x$

Step 5

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

$4 \int \frac{1}{\sqrt{x}} d x + \int \frac{3}{x} d x + \int - 2 e^{- x} d x$

Step 6

Apply basic rules of exponents.

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Step 6.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x}$ as $x^{\frac{1}{2}}$ .

$4 \int \frac{1}{x^{\frac{1}{2}}} d x + \int \frac{3}{x} d x + \int - 2 e^{- x} d x$

Step 6.2

Move $x^{\frac{1}{2}}$ out of the denominator by raising it to the $- 1$ power.

$4 \int {(x^{\frac{1}{2}})}^{- 1} d x + \int \frac{3}{x} d x + \int - 2 e^{- x} d x$

Step 6.3

Multiply the exponents in ${(x^{\frac{1}{2}})}^{- 1}$ .

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Step 6.3.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$4 \int x^{\frac{1}{2} \cdot - 1} d x + \int \frac{3}{x} d x + \int - 2 e^{- x} d x$

Step 6.3.2

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

$4 \int x^{\frac{- 1}{2}} d x + \int \frac{3}{x} d x + \int - 2 e^{- x} d x$

Step 6.3.3

Move the negative in front of the fraction.

$4 \int x^{- \frac{1}{2}} d x + \int \frac{3}{x} d x + \int - 2 e^{- x} d x$

Step 7

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

$4 (2 x^{\frac{1}{2}} + C) + \int \frac{3}{x} d x + \int - 2 e^{- x} d x$

Step 8

Since $3$ is constant with respect to $x$ , move $3$ out of the integral.

$4 (2 x^{\frac{1}{2}} + C) + 3 \int \frac{1}{x} d x + \int - 2 e^{- x} d x$

Step 9

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

$4 (2 x^{\frac{1}{2}} + C) + 3 (\ln (| x |) + C) + \int - 2 e^{- x} d x$

Step 10

Since $- 2$ is constant with respect to $x$ , move $- 2$ out of the integral.

$4 (2 x^{\frac{1}{2}} + C) + 3 (\ln (| x |) + C) - 2 \int e^{- x} d x$

Step 11

Let $u = - x$ . Then $d u = - d x$ , so $- d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 11.1

Let $u = - x$ . Find $\frac{d u}{d x}$ .

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Step 11.1.1

Differentiate $- x$ .

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

Step 11.1.2

Since $- 1$ is constant with respect to $x$ , the derivative of $- x$ with respect to $x$ is $- \frac{d}{d x} [x]$ .

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

Step 11.1.3

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

$- 1 \cdot 1$

Step 11.1.4

Multiply $- 1$ by $1$ .

$- 1$

Step 11.2

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

$4 (2 x^{\frac{1}{2}} + C) + 3 (\ln (| x |) + C) - 2 \int - e^{u} d u$

Step 12

Since $- 1$ is constant with respect to $u$ , move $- 1$ out of the integral.

$4 (2 x^{\frac{1}{2}} + C) + 3 (\ln (| x |) + C) - 2 (- \int e^{u} d u)$

Step 13

Multiply $- 1$ by $- 2$ .

$4 (2 x^{\frac{1}{2}} + C) + 3 (\ln (| x |) + C) + 2 \int e^{u} d u$

Step 14

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

$4 (2 x^{\frac{1}{2}} + C) + 3 (\ln (| x |) + C) + 2 (e^{u} + C)$

Step 15

Simplify.

$8 x^{\frac{1}{2}} + 3 \ln (| x |) + 2 e^{u} + C$

Step 16

Replace all occurrences of $u$ with $- x$ .

$8 x^{\frac{1}{2}} + 3 \ln (| x |) + 2 e^{- x} + C$

Step 17

The answer is the antiderivative of the function $f (x) = \frac{4}{\sqrt{x}} + \frac{3}{x} - 2 e^{- x}$ .

$F (x) =$ $8 x^{\frac{1}{2}} + 3 \ln (| x |) + 2 e^{- x} + C$