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

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

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

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

3 (- 2 ln (2) + 3)

Decimal Notation

4.84111 \dots

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

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

Expand

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

= 3 \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

= 3 (\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

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

\frac{d}{d x} (2 x + 7 + \frac{450}{x})

f (x) = \frac{2}{3} (x^{2} + 1)^{\frac{3}{2}}

y^{^{''}} + y = 0, y (\frac{π}{3}) = 0, y^{^{'}} (\frac{π}{3}) = 6

\int \frac{x + 1}{x ^{2}} d x

\int \frac{1}{x ^{2} + 5 x + 6} d x

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