integral from 0 to 4 of (8t)/((t-5)^2)

1) Let's compute the integral:

\int_{0}^{4} \frac{8 t}{{(t - 5)}^{2}} d t

2) You can decompose into partial fractions:

\frac{8 t}{{(t - 5)}^{2}} = \frac{A}{t - 5} + \frac{B}{{(t - 5)}^{2}}

To find the remaining unknowns, you multiply both sides by the denominator:

{(t - 5)}^{2}

which yields:

8 t = A \cdot (t - 5) + B

ⓘ Expand by multiplying each term inside the parentheses:

A \cdot (t - 5)

A t + A \cdot (-5)

8 t = A t + A \cdot (-5) + B

ⓘ The minus sign is placed in front of the product:

A \cdot (-5)

- 5 A

8 t = A t - 5 A + B

3) By identifying the different terms, you obtain the system of equations:

[1]:

8 = A

[2]:

0 = - 5 A + B

Solution found [1]:

A = 8

4) Let's solve for B:

0 = - 5 A + B

[2] where

A = 8

ⓘ Multiplications are done before additions and subtractions:

5 \cdot 8

40

- 40 + B = 0

ⓘ We add

40

to both sides

Solution obtained [2]:

B = 40

5) Finally, substitute the unknowns back into the initial partial fraction decomposition:

\frac{8 t}{{(t - 5)}^{2}} = \frac{8}{t - 5} + \frac{40}{{(t - 5)}^{2}}

6) Let's integrate:

\int (\frac{8}{t - 5} + \frac{40}{{(t - 5)}^{2}}) d t

ⅰ

\int (\frac{8}{t - 5} + \frac{40}{{(t - 5)}^{2}}) d t

\int \frac{8}{t - 5} d t + \int \frac{40}{{(t - 5)}^{2}} d t

\int \frac{8}{t - 5} d t + \int \frac{40}{{(t - 5)}^{2}} d t

Let

u = t - 5

. Then

du = dt

, so

dt = du

\int \frac{8}{t - 5} d t = 8 \cdot \int \frac{1}{u} d u

We use

\int \frac{1}{u} d u = \ln | u |

. The absolute value covers both signs of the denominator, on any interval where

t - 5 \neq 0

Substitute

u

t - 5

back into the antiderivative

Let

u = t - 5

. Then

du = dt

, so

dt = du

\int \frac{40}{{(t - 5)}^{2}} d t = 40 \cdot \int u^{-2} d u

Apply the power rule:

\int u^{-2} d u = - u^{-1}

Substitute

u

t - 5

back into the antiderivative

The antiderivative is valid on each interval where

t - 5 \neq 0

ⓘ The derivative of

\frac{40}{- t + 5}

\frac{40}{{(t - 5)}^{2}}

, so

\frac{40}{- t + 5}

is an antiderivative of

\frac{40}{{(t - 5)}^{2}}

\int \frac{40}{{(t - 5)}^{2}} d t

\frac{40}{- t + 5}

ⅰ

\int \frac{8 t}{{(t - 5)}^{2}} d t

8 \cdot \ln | t - 5 | + \frac{40}{- t + 5}

8 \cdot \ln | t - 5 | + \frac{40}{- t + 5}

We factor out

8

, the greatest common divisor of the coefficients:

8 \cdot \ln | t - 5 | + \frac{40}{- t + 5}

8 \cdot (\ln | t - 5 | + \frac{5}{- t + 5})

Factor found:

\ln | t - 5 | + \frac{5}{5 - t}

▷

8 \cdot (\ln | t - 5 | + \frac{5}{5 - t})

7) Let's evaluate over the interval

[0, 4]

8 \cdot (\ln | 4 - 5 | + \frac{5}{5 - 4}) - 8 \cdot (\ln | 0 - 5 | + \frac{5}{5 + 0})

ⓘ Start from

4

and move back

5

on the number line, going below zero:

4 - 5

-1

■The numerator and the denominator are computed before the division:

5 - 4

1

■Adding 0 changes nothing:

0 - 5

-5

5 + 0

5

■We simplify the fraction by dividing the numerator and the denominator by

5

, their greatest common divisor:

\frac{5}{5}

1

8 \cdot (\ln | -1 | + \frac{5}{1}) - 8 \cdot (\ln | -5 | + 1)

8 \cdot (\ln | -1 | + 5) - 8 \cdot (\ln | -5 | + 1)

ⓘ The expression inside the absolute value is nonpositive, so its absolute value equals its opposite:

| -1 |

1

| -5 |

5

■Expand by multiplying each term inside the parentheses:

8 \cdot (\ln (1) + 5)

8 \cdot \ln (1) + 40

- 8 \cdot (\ln (5) + 1)

- (8 \cdot \ln (5) + 8)

8 \cdot \ln (1) + 40 - (8 \cdot \ln (5) + 8)

ⓘ ln(1) = 0, since exp(0) = 1:

\ln (1)

0

■A minus sign before a parenthesis changes the sign of each of its terms:

- (8 \cdot \ln (5) + 8)

- 8 \cdot \ln (5) - 8

8 \cdot 0

+ 40

- 8 \cdot \ln (5)

- 8

ⓘ A product with a zero factor is zero:

8 \cdot 0

0

■Start from

40

and move back

8

on the number line:

40 - 8

32

0 + 32 - 8 \cdot \ln (5)

ⓘ Adding 0 changes nothing:

0 + 32

32

▶Integral:

32 - 8 \cdot \ln (5)

8) The numeric value of:

32 - 8 \cdot \ln (5)

equals:

32 - 8 \cdot \approx 1.6094

32 - \approx 12.8755

▷

\approx 19.1245