What is the integral from 0 to 2 of (8t)/((t-3)^2) ?

The integral from 0 to 2 of (8t)/((t-3)^2) is 8(-ln(3)+2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 2 of (8t)/((t-3)^2)

\int_{0}^{2} \frac{8 t}{( t - 3 ) ^{2}} d t

8 (- ln (3) + 2)

Decimal Notation

7.21110 \dots

Solution steps

\int_{0}^{2} \frac{8 t}{( t - 3 ) ^{2}} d t

Take the constant out:

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

= 8 \cdot \int_{0}^{2} \frac{t}{( t - 3 ) ^{2}} d t

Apply u-substitution

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

= 8 \cdot \int_{- 3}^{- 1} \frac{u + 3}{u ^{2}} d u

Expand

\frac{u + 3}{u ^{2}} : \frac{1}{u} + \frac{3}{u ^{2}}

= 8 \cdot \int_{- 3}^{- 1} \frac{1}{u} + \frac{3}{u ^{2}} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 8 (\int_{- 3}^{- 1} \frac{1}{u} d u + \int_{- 3}^{- 1} \frac{3}{u ^{2}} d u)

\int_{- 3}^{- 1} \frac{1}{u} d u = - ln (3)

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

= 8 (- ln (3) + 2)

\frac{d}{d x} (\frac{x - 1}{3})

\frac{6 x}{x ^{2} + 1} (1.3)

\frac{7 x}{( x ^{3} - 2 ) ^{2}}

\int - \frac{cos ( 2 x )}{4} d x

\frac{d y}{d x} = \frac{2}{3} y

What is the integral from 0 to 2 of (8t)/((t-3)^2) ?
The integral from 0 to 2 of (8t)/((t-3)^2) is 8(-ln(3)+2)