What is the integral from 6 to 8 of (52)/((x-6)^3) ?

The integral from 6 to 8 of (52)/((x-6)^3) is diverges

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 6 to 8 of (52)/((x-6)^3)

\int_{6}^{8} \frac{52}{( x - 6 ) ^{3}} d x

diverges

Solution steps

\int_{6}^{8} \frac{52}{( x - 6 ) ^{3}} d x

Take the constant out:

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

= 52 \cdot \int_{6}^{8} \frac{1}{( x - 6 ) ^{3}} d x

Apply u-substitution

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

= 52 \cdot \int_{0}^{2} \frac{1}{u ^{3}} d u

Apply the Power Rule

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

= 52 [- \frac{1}{2 u ^{2}}]_{0}^{2}

Compute the boundaries:

diverges

= diverges

\int \frac{18 x}{1 - 9 x ^{2}} d x

\int 3 x (6 - x^{2}) d x

\int (- 1)^{n} x^{n} d x

\int y^{3} (2 y^{2} - 3) d y

\frac{\partial}{\partial x} (2 y + y e^{x})

What is the integral from 6 to 8 of (52)/((x-6)^3) ?
The integral from 6 to 8 of (52)/((x-6)^3) is diverges