What is the integral from 0 to 3 of 1/((x-3)^2) ?

The integral from 0 to 3 of 1/((x-3)^2) is diverges

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 3 of 1/((x-3)^2)

\int_{0}^{3} \frac{1}{( x - 3 ) ^{2}} d x

diverges

Solution steps

\int_{0}^{3} \frac{1}{( x - 3 ) ^{2}} d x

Apply u-substitution

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

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

Apply the Power Rule

: [- \frac{1}{u}]_{- 3}^{0}

= [- \frac{1}{u}]_{- 3}^{0}

Compute the boundaries:

diverges

= diverges

\int_{0}^{1} x^{4} x^{3} d x

\int_{3}^{\infty} t e^{- s t} d t

\int_{- 4}^{4} x^{2} - 3 d x

\int_{- 1}^{2} 6 x (x^{2} - 1)^{2} d x

\int_{1}^{6} \frac{1}{25 + ( x + 4 ) ^{2}} d x

What is the integral from 0 to 3 of 1/((x-3)^2) ?
The integral from 0 to 3 of 1/((x-3)^2) is diverges