What is the integral from 0 to 3 of (6t)/((t-4)^2) ?

The integral from 0 to 3 of (6t)/((t-4)^2) is 6(-2ln(2)+3)

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 (6t)/((t-4)^2)

\int_{0}^{3} \frac{6 t}{( t - 4 ) ^{2}} d t

6 (- 2 ln (2) + 3)

Decimal Notation

9.68223 \dots

Solution steps

\int_{0}^{3} \frac{6 t}{( t - 4 ) ^{2}} d t

Take the constant out:

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

= 6 \cdot \int_{0}^{3} \frac{t}{( t - 4 ) ^{2}} d t

Apply u-substitution

: \int_{- 4}^{- 1} \frac{u + 4}{u ^{2}} d u

= 6 \cdot \int_{- 4}^{- 1} \frac{u + 4}{u ^{2}} d u

Expand

\frac{u + 4}{u ^{2}} : \frac{1}{u} + \frac{4}{u ^{2}}

= 6 \cdot \int_{- 4}^{- 1} \frac{1}{u} + \frac{4}{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

= 6 (\int_{- 4}^{- 1} \frac{1}{u} d u + \int_{- 4}^{- 1} \frac{4}{u ^{2}} d u)

\int_{- 4}^{- 1} \frac{1}{u} d u = - 2 ln (2)

\int_{- 4}^{- 1} \frac{4}{u ^{2}} d u = 3

= 6 (- 2 ln (2) + 3)

\int \frac{2 x e ^{x}}{( x + 1 ) ^{2}} d x

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

sin (x^{2}), 0

y^{^{''}} + y^{^{'}} - 2 y = sin (x)

\frac{d y}{d x} - y = e^{2 x} y^{3}

What is the integral from 0 to 3 of (6t)/((t-4)^2) ?
The integral from 0 to 3 of (6t)/((t-4)^2) is 6(-2ln(2)+3)