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

The integral from 5 to x^2 of 1/(t^4) is -1/(3x^6)+1/375

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integral from 5 to x^2 of 1/(t^4)

\int_{5}^{x^{2}} \frac{1}{t ^{4}} d t

- \frac{1}{3 x ^{6}} + \frac{1}{375}

Solution steps

\int_{5}^{x^{2}} \frac{1}{t ^{4}} d t

Apply the Power Rule

: [- \frac{1}{3 t ^{3}}]_{5}^{x^{2}}

= [- \frac{1}{3 t ^{3}}]_{5}^{x^{2}}

Compute the boundaries:

- \frac{1}{3 x ^{6}} + \frac{1}{375}

= - \frac{1}{3 x ^{6}} + \frac{1}{375}

\int_{0}^{x} 3 t^{2} d t

\int_{0}^{\frac{7}{3}} 3 t - 7 d t

\int_{0}^{2} k x (2 - x)^{2} d x

\int_{3}^{10} 5 d x

\int_{0}^{t} \frac{6}{1 + 6 t ^{2}} d t

What is the integral from 5 to x^2 of 1/(t^4) ?
The integral from 5 to x^2 of 1/(t^4) is -1/(3x^6)+1/375