Calculus Examples

$\frac{3 z^{2} + 12 z - 9}{z^{4}}$

Step 1

Move $z^{4}$ out of the denominator by raising it to the $- 1$ power.

$\int (3 z^{2} + 12 z - 9) {(z^{4})}^{- 1} d z$

Step 2

Multiply the exponents in ${(z^{4})}^{- 1}$ .

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Step 2.1

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

$\int (3 z^{2} + 12 z - 9) z^{4 \cdot - 1} d z$

Step 2.2

Multiply $4$ by $- 1$ .

$\int (3 z^{2} + 12 z - 9) z^{- 4} d z$

Step 3

Expand $(3 z^{2} + 12 z - 9) z^{- 4}$ .

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Step 3.1

Apply the distributive property.

$\int (3 z^{2} + 12 z) z^{- 4} - 9 z^{- 4} d z$

Step 3.2

Apply the distributive property.

$\int 3 z^{2} z^{- 4} + 12 z \cdot z^{- 4} - 9 z^{- 4} d z$

Step 3.3

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\int 3 z^{2 - 4} + 12 z \cdot z^{- 4} - 9 z^{- 4} d z$

Step 3.4

Subtract $4$ from $2$ .

$\int 3 z^{- 2} + 12 z \cdot z^{- 4} - 9 z^{- 4} d z$

Step 3.5

Raise $z$ to the power of $1$ .

$\int 3 z^{- 2} + 12 (z^{1} z^{- 4}) - 9 z^{- 4} d z$

Step 3.6

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\int 3 z^{- 2} + 12 z^{1 - 4} - 9 z^{- 4} d z$

Step 3.7

Subtract $4$ from $1$ .

$\int 3 z^{- 2} + 12 z^{- 3} - 9 z^{- 4} d z$

Step 4

Split the single integral into multiple integrals.

$\int 3 z^{- 2} d z + \int 12 z^{- 3} d z + \int - 9 z^{- 4} d z$

Step 5

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

$3 \int z^{- 2} d z + \int 12 z^{- 3} d z + \int - 9 z^{- 4} d z$

Step 6

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

$3 (- z^{- 1} + C) + \int 12 z^{- 3} d z + \int - 9 z^{- 4} d z$

Step 7

Since $12$ is constant with respect to $z$ , move $12$ out of the integral.

$3 (- z^{- 1} + C) + 12 \int z^{- 3} d z + \int - 9 z^{- 4} d z$

Step 8

By the Power Rule, the integral of $z^{- 3}$ with respect to $z$ is $- \frac{1}{2} z^{- 2}$ .

$3 (- z^{- 1} + C) + 12 (- \frac{1}{2} z^{- 2} + C) + \int - 9 z^{- 4} d z$

Step 9

Simplify.

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Step 9.1

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

$3 (- z^{- 1} + C) + 12 (- \frac{z^{- 2}}{2} + C) + \int - 9 z^{- 4} d z$

Step 9.2

Move $z^{- 2}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$3 (- z^{- 1} + C) + 12 (- \frac{1}{2 z^{2}} + C) + \int - 9 z^{- 4} d z$

Step 10

Since $- 9$ is constant with respect to $z$ , move $- 9$ out of the integral.

$3 (- z^{- 1} + C) + 12 (- \frac{1}{2 z^{2}} + C) - 9 \int z^{- 4} d z$

Step 11

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

$3 (- z^{- 1} + C) + 12 (- \frac{1}{2 z^{2}} + C) - 9 (- \frac{1}{3} z^{- 3} + C)$

Step 12

Simplify.

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Step 12.1

Simplify.

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Step 12.1.1

Combine $z^{- 3}$ and $\frac{1}{3}$ .

$3 (- z^{- 1} + C) + 12 (- \frac{1}{2 z^{2}} + C) - 9 (- \frac{z^{- 3}}{3} + C)$

Step 12.1.2

Move $z^{- 3}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$3 (- z^{- 1} + C) + 12 (- \frac{1}{2 z^{2}} + C) - 9 (- \frac{1}{3 z^{3}} + C)$

Step 12.2

Simplify.

$- \frac{3}{z} - \frac{6}{z^{2}} - 9 (- \frac{1}{3 z^{3}}) + C$

Step 12.3

Simplify.

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Step 12.3.1

Multiply $- 1$ by $- 9$ .

$- \frac{3}{z} - \frac{6}{z^{2}} + 9 \frac{1}{3 z^{3}} + C$

Step 12.3.2

Combine $9$ and $\frac{1}{3 z^{3}}$ .

$- \frac{3}{z} - \frac{6}{z^{2}} + \frac{9}{3 z^{3}} + C$

Step 12.3.3

Cancel the common factor of $9$ and $3$ .

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Step 12.3.3.1

Factor $3$ out of $9$ .

$- \frac{3}{z} - \frac{6}{z^{2}} + \frac{3 \cdot 3}{3 z^{3}} + C$

Step 12.3.3.2

Cancel the common factors.

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Step 12.3.3.2.1

Factor $3$ out of $3 z^{3}$ .

$- \frac{3}{z} - \frac{6}{z^{2}} + \frac{3 \cdot 3}{3 (z^{3})} + C$

Step 12.3.3.2.2

Cancel the common factor.

$- \frac{3}{z} - \frac{6}{z^{2}} + \frac{3 \cdot 3}{3 z^{3}} + C$

Step 12.3.3.2.3

Rewrite the expression.

$- \frac{3}{z} - \frac{6}{z^{2}} + \frac{3}{z^{3}} + C$