Calculus Examples

$\frac{x^{5}}{x^{2} + 1}$

Step 1

Write $\frac{x^{5}}{x^{2} + 1}$ as a function.

$f (x) = \frac{x^{5}}{x^{2} + 1}$

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.

$F (x) = \int \frac{x^{5}}{x^{2} + 1} d x$

Step 4

Divide $x^{5}$ by $x^{2} + 1$ .

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Step 4.1

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$x^{2}$

$0 x$

$1$

$x^{5}$

$0 x^{4}$

$0 x^{3}$

$0 x^{2}$

$0 x$

$0$

Step 4.2

Divide the highest order term in the dividend $x^{5}$ by the highest order term in divisor $x^{2}$ .

$x^{3}$

$x^{2}$

$0 x$

$1$

$x^{5}$

$0 x^{4}$

$0 x^{3}$

$0 x^{2}$

$0 x$

$0$

Step 4.3

Multiply the new quotient term by the divisor.

$x^{3}$

$x^{2}$

$0 x$

$1$

$x^{5}$

$0 x^{4}$

$0 x^{3}$

$0 x^{2}$

$0 x$

$0$

$x^{5}$

$0$

$x^{3}$

Step 4.4

The expression needs to be subtracted from the dividend, so change all the signs in $x^{5} + 0 + x^{3}$

$x^{3}$

$x^{2}$

$0 x$

$1$

$x^{5}$

$0 x^{4}$

$0 x^{3}$

$0 x^{2}$

$0 x$

$0$

$x^{5}$

$0$

$x^{3}$

Step 4.5

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

$x^{3}$

$x^{2}$

$0 x$

$1$

$x^{5}$

$0 x^{4}$

$0 x^{3}$

$0 x^{2}$

$0 x$

$0$

$x^{5}$

$0$

$x^{3}$

Step 4.6

Pull the next term from the original dividend down into the current dividend.

$x^{3}$

$x^{2}$

$0 x$

$1$

$x^{5}$

$0 x^{4}$

$0 x^{3}$

$0 x^{2}$

$0 x$

$0$

$x^{5}$

$0$

$x^{3}$

$0 x^{2}$

$0 x$

Step 4.7

Divide the highest order term in the dividend $- x^{3}$ by the highest order term in divisor $x^{2}$ .

$x^{3}$

$0 x^{2}$

$x$

$x^{2}$

$0 x$

$1$

$x^{5}$

$0 x^{4}$

$0 x^{3}$

$0 x^{2}$

$0 x$

$0$

$x^{5}$

$0$

$x^{3}$

$0 x^{2}$

$0 x$

Step 4.8

Multiply the new quotient term by the divisor.

$x^{3}$

$0 x^{2}$

$x$

$x^{2}$

$0 x$

$1$

$x^{5}$

$0 x^{4}$

$0 x^{3}$

$0 x^{2}$

$0 x$

$0$

$x^{5}$

$0$

$x^{3}$

$0 x^{2}$

$0 x$

$x^{3}$

$0$

$x$

Step 4.9

The expression needs to be subtracted from the dividend, so change all the signs in $- x^{3} + 0 - x$

$x^{3}$

$0 x^{2}$

$x$

$x^{2}$

$0 x$

$1$

$x^{5}$

$0 x^{4}$

$0 x^{3}$

$0 x^{2}$

$0 x$

$0$

$x^{5}$

$0$

$x^{3}$

$0 x^{2}$

$0 x$

$x^{3}$

$0$

$x$

Step 4.10

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

$x^{3}$

$0 x^{2}$

$x$

$x^{2}$

$0 x$

$1$

$x^{5}$

$0 x^{4}$

$0 x^{3}$

$0 x^{2}$

$0 x$

$0$

$x^{5}$

$0$

$x^{3}$

$0 x^{2}$

$0 x$

$x^{3}$

$0$

$x$

Step 4.11

Pull the next term from the original dividend down into the current dividend.

$x^{3}$

$0 x^{2}$

$x$

$x^{2}$

$0 x$

$1$

$x^{5}$

$0 x^{4}$

$0 x^{3}$

$0 x^{2}$

$0 x$

$0$

$x^{5}$

$0$

$x^{3}$

$0 x^{2}$

$0 x$

$x^{3}$

$0$

$x$

$0$

Step 4.12

The final answer is the quotient plus the remainder over the divisor.

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

Step 5

Split the single integral into multiple integrals.

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

Step 6

By the Power Rule, the integral of $x^{3}$ with respect to $x$ is $\frac{1}{4} x^{4}$ .

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

Step 7

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

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

Step 8

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

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

Step 9

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

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Step 9.1

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

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Step 9.1.1

Differentiate $x^{2} + 1$ .

$\frac{d}{d x} [x^{2} + 1]$

Step 9.1.2

By the Sum Rule, the derivative of $x^{2} + 1$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [1]$ .

$\frac{d}{d x} [x^{2}] + \frac{d}{d x} [1]$

Step 9.1.3

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$2 x + \frac{d}{d x} [1]$

Step 9.1.4

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

$2 x + 0$

Step 9.1.5

Add $2 x$ and $0$ .

$2 x$

Step 9.2

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

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

Step 10

Simplify.

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Step 10.1

Combine $\frac{1}{2}$ and $x^{2}$ .

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

Step 10.2

Multiply $\frac{1}{u}$ by $\frac{1}{2}$ .

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

Step 10.3

Move $2$ to the left of $u$ .

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

Step 11

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

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

Step 12

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

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

Step 13

Simplify.

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

Step 14

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

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

Step 15

The answer is the antiderivative of the function $f (x) = \frac{x^{5}}{x^{2} + 1}$ .

$F (x) =$ $\frac{1}{4} x^{4} - \frac{1}{2} x^{2} + \frac{1}{2} \ln (| x^{2} + 1 |) + C$