What is the integral from 0 to 5 of (2t)/((t-6)^2) ?

The integral from 0 to 5 of (2t)/((t-6)^2) is 2(-ln(6)+5)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 5 of (2t)/((t-6)^2)

\int_{0}^{5} \frac{2 t}{( t - 6 ) ^{2}} d t

2 (- ln (6) + 5)

Decimal Notation

6.41648 \dots

Solution steps

\int_{0}^{5} \frac{2 t}{( t - 6 ) ^{2}} d t

Take the constant out:

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

= 2 \cdot \int_{0}^{5} \frac{t}{( t - 6 ) ^{2}} d t

Apply u-substitution

: \int_{- 6}^{- 1} \frac{u + 6}{u ^{2}} d u

= 2 \cdot \int_{- 6}^{- 1} \frac{u + 6}{u ^{2}} d u

Expand

\frac{u + 6}{u ^{2}} : \frac{1}{u} + \frac{6}{u ^{2}}

= 2 \cdot \int_{- 6}^{- 1} \frac{1}{u} + \frac{6}{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

= 2 (\int_{- 6}^{- 1} \frac{1}{u} d u + \int_{- 6}^{- 1} \frac{6}{u ^{2}} d u)

\int_{- 6}^{- 1} \frac{1}{u} d u = - ln (6)

\int_{- 6}^{- 1} \frac{6}{u ^{2}} d u = 5

= 2 (- ln (6) + 5)

\int \frac{x + 2}{x ^{2} + 6 x + 10} d x

x y^{^{'}} + 3 y = \frac{( e ^{x} )}{x}

cos^{3} (x) + sin^{3} (x)

\int sin^{2} (x co) s^{3} x d x

\frac{\partial}{\partial y} (5 x sin (y))

What is the integral from 0 to 5 of (2t)/((t-6)^2) ?
The integral from 0 to 5 of (2t)/((t-6)^2) is 2(-ln(6)+5)