What is the integral from 0 to 1 of (69)/(x^5) ?

The integral from 0 to 1 of (69)/(x^5) is diverges

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integral from 0 to 1 of (69)/(x^5)

\int_{0}^{1} \frac{69}{x ^{5}} d x

diverges

Solution steps

\int_{0}^{1} \frac{69}{x ^{5}} d x

Take the constant out:

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

= 69 \cdot \int_{0}^{1} \frac{1}{x ^{5}} d x

Apply the Power Rule

: [- \frac{1}{4 x ^{4}}]_{0}^{1}

= 69 [- \frac{1}{4 x ^{4}}]_{0}^{1}

Compute the boundaries:

diverges

= diverges

5 t sin (π t)

y = e^{2 x}, y = e^{3 x}, x = 1

x \to - 6 lim (- 5 x - 6)

f (x) = \frac{2 x}{x ^{2} - 7}, at x = 3

\int \frac{sin ( 3 x )}{3 cos ( 3 x )} d x

What is the integral from 0 to 1 of (69)/(x^5) ?
The integral from 0 to 1 of (69)/(x^5) is diverges