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

The integral from 1 to x^2 of 1/(t^5) is -1/(4x^8)+1/4

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

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

- \frac{1}{4 x ^{8}} + \frac{1}{4}

Solution steps

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

Apply the Power Rule

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

Compute the boundaries:

- \frac{1}{4 x ^{8}} + \frac{1}{4}

= - \frac{1}{4 x ^{8}} + \frac{1}{4}

What is the integral from 1 to x^2 of 1/(t^5) ?
The integral from 1 to x^2 of 1/(t^5) is -1/(4x^8)+1/4