What is the integral from 6 to 8 of (76)/((x-6)^3) ?

The integral from 6 to 8 of (76)/((x-6)^3) is diverges

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integral from 6 to 8 of (76)/((x-6)^3)

\int_{6}^{8} \frac{76}{( x - 6 ) ^{3}} d x

diverges

Solution steps

\int_{6}^{8} \frac{76}{( x - 6 ) ^{3}} d x

Take the constant out:

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

= 76 \cdot \int_{6}^{8} \frac{1}{( x - 6 ) ^{3}} d x

Apply u-substitution

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

= 76 \cdot \int_{0}^{2} \frac{1}{u ^{3}} d u

Apply the Power Rule

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

= 76 [- \frac{1}{2 u ^{2}}]_{0}^{2}

Compute the boundaries:

diverges

= diverges

\frac{d y}{d x}, x^{2} y = 1

\int \frac{2 x}{λ ^{2}} d x

x \to 0 lim (\frac{e ^{4 x} - 1}{2 x ^{2} + 2 x})

\int sin (x) sin (n) x d x

n = 1 \sum \infty \frac{4}{n ( n + 6 )}

What is the integral from 6 to 8 of (76)/((x-6)^3) ?
The integral from 6 to 8 of (76)/((x-6)^3) is diverges