What is the integral from 1 to x of 3/(x^4) ?

The integral from 1 to x of 3/(x^4) is -1/(x^3)+1

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integral from 1 to x of 3/(x^4)

\int_{1}^{x} \frac{3}{x ^{4}} d x

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

Solution steps

\int_{1}^{x} \frac{3}{x ^{4}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 \cdot \int_{1}^{x} \frac{1}{x ^{4}} d x

Apply the Power Rule

= 3 [- \frac{1}{3 x ^{3}}]_{1}^{x}

Compute the boundaries:

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

= 3 (- \frac{1}{3 x ^{3}} + \frac{1}{3})

Simplify

= - \frac{1}{x ^{3}} + 1

\int_{- 8}^{- 3} - x^{2} - 3 x d x

\int_{0}^{6} (\frac{1}{4 - x}) d x

\int_{- 1}^{1} e^{- x} - 1 d x

\int_{1}^{3} \frac{1}{x + x ^{2}} d x

\int_{1}^{1} 2 x d x

What is the integral from 1 to x of 3/(x^4) ?
The integral from 1 to x of 3/(x^4) is -1/(x^3)+1