What is the integral from 6 to 8 of (70)/((x-6)^3) ?

The integral from 6 to 8 of (70)/((x-6)^3) is diverges

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integral from 6 to 8 of (70)/((x-6)^3)

\int_{6}^{8} \frac{70}{( x - 6 ) ^{3}} d x

diverges

Solution steps

\int_{6}^{8} \frac{70}{( x - 6 ) ^{3}} d x

Take the constant out:

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

= 70 \cdot \int_{6}^{8} \frac{1}{( x - 6 ) ^{3}} d x

Apply u-substitution

: \int_{0}^{2} \frac{1}{u ^{3}} d u

= 70 \cdot \int_{0}^{2} \frac{1}{u ^{3}} d u

Apply the Power Rule

: [- \frac{1}{2 u ^{2}}]_{0}^{2}

= 70 [- \frac{1}{2 u ^{2}}]_{0}^{2}

Compute the boundaries:

diverges

= diverges

1 - 3 x^{2}

f^{^{'}} (x) = f (x) (1 - f (x))

\int \frac{( 7 sin ( x ) + 4 sec ( x ) )}{tan ( x )} d x

f (x) = 200 + \frac{7}{x} + \frac{x ^{2}}{7}

y^{^{'}} = x (9 - y)

What is the integral from 6 to 8 of (70)/((x-6)^3) ?
The integral from 6 to 8 of (70)/((x-6)^3) is diverges