Find the Integral 1/(x^4)

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

Apply basic rules of exponents.

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

$\int {(x^{4})}^{- 1} d x$

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

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

$\int x^{4 \cdot - 1} d x$

Multiply $4$ by $- 1$ .

$\int x^{- 4} d x$

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

$- \frac{1}{3} x^{- 3} + C$

Simplify the answer.

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Rewrite $- \frac{1}{3} x^{- 3} + C$ as $- \frac{1}{3} \cdot \frac{1}{x^{3}} + C$ .

$- \frac{1}{3} \cdot \frac{1}{x^{3}} + C$

Simplify.

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Multiply $\frac{1}{x^{3}}$ and $\frac{1}{3}$ .

$- \frac{1}{x^{3} \cdot 3} + C$

Move $3$ to the left of $x^{3}$ .

$- \frac{1}{3 x^{3}} + C$

Find the Integral 1/(x^4)

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