Calculus Examples

$\int \frac{5 x}{4 x + 4 x^{2}} d x$

Step 1

Since $5$ is constant with respect to $x$ , move $5$ out of the integral.

$5 \int \frac{x}{4 x + 4 x^{2}} d x$

Step 2

Write the fraction using partial fraction decomposition.

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Step 2.1

Decompose the fraction and multiply through by the common denominator.

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Step 2.1.1

Factor the fraction.

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Step 2.1.1.1

Factor $4 x$ out of $4 x + 4 x^{2}$ .

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Step 2.1.1.1.1

Factor $4 x$ out of $4 x$ .

$\frac{x}{4 x (1) + 4 x^{2}}$

Step 2.1.1.1.2

Factor $4 x$ out of $4 x^{2}$ .

$\frac{x}{4 x (1) + 4 x (x)}$

Step 2.1.1.1.3

Factor $4 x$ out of $4 x (1) + 4 x (x)$ .

$\frac{x}{4 x (1 + x)}$

Step 2.1.1.2

Reduce the expression $\frac{x}{4 x (1 + x)}$ by cancelling the common factors.

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Step 2.1.1.2.1

Cancel the common factor.

$\frac{x}{4 x (1 + x)}$

Step 2.1.1.2.2

Rewrite the expression.

$\frac{1}{4 (1 + x)}$

Step 2.1.2

For each factor in the denominator, create a new fraction using the factor as the denominator, and an unknown value as the numerator. Since the factor in the denominator is linear, put a single variable in its place $A$ .

$\frac{A}{1 + x}$

Step 2.1.3

Multiply each fraction in the equation by the denominator of the original expression. In this case, the denominator is $4 (1 + x)$ .

$\frac{1 (4 (1 + x))}{4 (1 + x)} = \frac{(A) (4 (1 + x))}{1 + x}$

Step 2.1.4

Cancel the common factor of $4$ .

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Step 2.1.4.1

Cancel the common factor.

$\frac{1 (4 (1 + x))}{4 (1 + x)} = \frac{(A) (4 (1 + x))}{1 + x}$

Step 2.1.4.2

Rewrite the expression.

$\frac{1 (1 + x)}{1 + x} = \frac{(A) (4 (1 + x))}{1 + x}$

Step 2.1.5

Cancel the common factor of $1 + x$ .

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Step 2.1.5.1

Cancel the common factor.

$\frac{1 (1 + x)}{1 + x} = \frac{(A) (4 (1 + x))}{1 + x}$

Step 2.1.5.2

Rewrite the expression.

$1 = \frac{(A) (4 (1 + x))}{1 + x}$

Step 2.1.6

Cancel the common factor of $1 + x$ .

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Step 2.1.6.1

Cancel the common factor.

$1 = \frac{A (4 (1 + x))}{1 + x}$

Step 2.1.6.2

Divide $(A) \cdot (4)$ by $1$ .

$1 = (A) \cdot (4)$

Step 2.1.7

Move $4$ to the left of $A$ .

$1 = 4 A$

Step 2.2

Solve the system of equations.

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Step 2.2.1

Rewrite the equation as $4 A = 1$ .

$4 A = 1$

Step 2.2.2

Divide each term in $4 A = 1$ by $4$ and simplify.

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Step 2.2.2.1

Divide each term in $4 A = 1$ by $4$ .

$\frac{4 A}{4} = \frac{1}{4}$

Step 2.2.2.2

Simplify the left side.

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Step 2.2.2.2.1

Cancel the common factor of $4$ .

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Step 2.2.2.2.1.1

Cancel the common factor.

$\frac{4 A}{4} = \frac{1}{4}$

Step 2.2.2.2.1.2

Divide $A$ by $1$ .

$A = \frac{1}{4}$

Step 2.3

Replace each of the partial fraction coefficients in $\frac{A}{1 + x}$ with the values found for and $A$ .

$\frac{\frac{1}{4}}{1 + x}$

Step 2.4

Simplify.

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Step 2.4.1

Multiply the numerator by the reciprocal of the denominator.

$\frac{1}{4} \cdot \frac{1}{1 + x}$

Step 2.4.2

Multiply $\frac{1}{4}$ by $\frac{1}{1 + x}$ .

$5 \int \frac{1}{4 (1 + x)} d x$

Step 3

Since $\frac{1}{4}$ is constant with respect to $x$ , move $\frac{1}{4}$ out of the integral.

$5 (\frac{1}{4} \int \frac{1}{1 + x} d x)$

Step 4

Combine $\frac{1}{4}$ and $5$ .

$\frac{5}{4} \int \frac{1}{1 + x} d x$

Step 5

Let $u = 1 + x$ . Then $d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 5.1

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

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Step 5.1.1

Differentiate $1 + x$ .

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

Step 5.1.2

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

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

Step 5.1.3

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

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

Step 5.1.4

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

$0 + 1$

Step 5.1.5

Add $0$ and $1$ .

$1$

Step 5.2

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

$\frac{5}{4} \int \frac{1}{u} d u$

Step 6

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

$\frac{5}{4} (\ln (| u |) + C)$

Step 7

Simplify.

$\frac{5}{4} \ln (| u |) + C$

Step 8

Replace all occurrences of $u$ with $1 + x$ .

$\frac{5}{4} \ln (| 1 + x |) + C$