What is the integral from 2 to t of 1/(5x^5) ?

The integral from 2 to t of 1/(5x^5) is 1/5 (-1/(4t^4)+1/64)

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integral from 2 to t of 1/(5x^5)

\int_{2}^{t} \frac{1}{5 x ^{5}} d x

\frac{1}{5} (- \frac{1}{4 t ^{4}} + \frac{1}{64})

Solution steps

\int_{2}^{t} \frac{1}{5 x ^{5}} d x

Take the constant out:

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

= \frac{1}{5} \cdot \int_{2}^{t} \frac{1}{x ^{5}} d x

Apply the Power Rule

: [- \frac{1}{4 x ^{4}}]_{2}^{t}

= \frac{1}{5} [- \frac{1}{4 x ^{4}}]_{2}^{t}

Compute the boundaries:

- \frac{1}{4 t ^{4}} + \frac{1}{64}

= \frac{1}{5} (- \frac{1}{4 t ^{4}} + \frac{1}{64})

\int \frac{1}{7 + 6 x - x ^{2}} d x

n = 1 \sum \infty (sin (1))^{n}

\int - 3 x d x

\frac{s - 2}{( s + 1 ) ( s - 3 )}

(\frac{1}{1 - \frac{x}{2}})^{3}, 0

What is the integral from 2 to t of 1/(5x^5) ?
The integral from 2 to t of 1/(5x^5) is 1/5 (-1/(4t^4)+1/64)