What is the integral from 0 to 3 of (7t)/((t-4)^2) ?

The integral from 0 to 3 of (7t)/((t-4)^2) is 7(-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 (7t)/((t-4)^2)

\int_{0}^{3} \frac{7 t}{( t - 4 ) ^{2}} d t

7 (- 2 ln (2) + 3)

Decimal Notation

11.29593 \dots

Solution steps

\int_{0}^{3} \frac{7 t}{( t - 4 ) ^{2}} d t

Take the constant out:

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

= 7 \cdot \int_{0}^{3} \frac{t}{( t - 4 ) ^{2}} d t

Apply u-substitution

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

= 7 \cdot \int_{- 4}^{- 1} \frac{u + 4}{u ^{2}} d u

Expand

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

= 7 \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

= 7 (\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

= 7 (- 2 ln (2) + 3)

7 x^{e} + e^{9 x}

f (x) = \frac{x ^{2}}{5 + x}

\int (a^{2} - x^{2})^{\frac{5}{2}} d x

\int 4 x^{3} - 10 x d x

\int \frac{4 t ^{3} - t ^{2} + 16 t}{t ^{2} + 4} d t

What is the integral from 0 to 3 of (7t)/((t-4)^2) ?
The integral from 0 to 3 of (7t)/((t-4)^2) is 7(-2ln(2)+3)