Solve \int (5)/((t^(2)-25)^(2))

Evaluate

\int \frac{5}{( t ^{2} - 25 ) ^{2}} d t

Rewrite the expression

\int 5 \times \frac{1}{( t ^{2} - 25 ) ^{2}} d t

Use the property of integral \int k f (x) d x = k \int f (x) d x

5 \times \int \frac{1}{( t ^{2} - 25 ) ^{2}} d t

Rewrite the fraction

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Evaluate

\frac{1}{( t ^{2} - 25 ) ^{2}}

Rewrite the expression

\frac{1}{( t - 5 ) ^{2} ( t + 5 ) ^{2}}

For each factor in the denominator,write a new fraction

\frac{?}{( t - 5 ) ^{2}} + \frac{?}{t - 5} + \frac{?}{( t + 5 ) ^{2}} + \frac{?}{t + 5}

Write the terms in the numerator

\frac{A}{( t - 5 ) ^{2}} + \frac{B}{t - 5} + \frac{C}{( t + 5 ) ^{2}} + \frac{D}{t + 5}

Set the sum of fractions equal to the original fraction

\frac{1}{( t - 5 ) ^{2} ( t + 5 ) ^{2}} = \frac{A}{( t - 5 ) ^{2}} + \frac{B}{t - 5} + \frac{C}{( t + 5 ) ^{2}} + \frac{D}{t + 5}

Multiply both sides

\frac{1}{( t - 5 ) ^{2} ( t + 5 ) ^{2}} \times (t - 5)^{2} (t + 5)^{2} = \frac{A}{( t - 5 ) ^{2}} \times (t - 5)^{2} (t + 5)^{2} + \frac{B}{t - 5} \times (t - 5)^{2} (t + 5)^{2} + \frac{C}{( t + 5 ) ^{2}} \times (t - 5)^{2} (t + 5)^{2} + \frac{D}{t + 5} \times (t - 5)^{2} (t + 5)^{2}

Simplify the expression

1 = (t^{2} + 10 t + 25) A + (t^{3} + 5 t^{2} - 25 t - 125) B + (t^{2} - 10 t + 25) C + (t^{3} - 5 t^{2} - 25 t + 125) D

Simplify the expression

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1 = t^{2} A + 10 t A + 25 A + t^{3} B + 5 t^{2} B - 25 tB - 125 B + t^{2} C - 10 tC + 25 C + t^{3} D - 5 t^{2} D - 25 t D + 125 D

Group the terms

1 = (B + D) t^{3} + (A + 5 B + C - 5 D) t^{2} + (10 A - 25 B - 10 C - 25 D) t + 25 A - 125 B + 25 C + 125 D

Equate the coefficients

⎩ ⎨ ⎧ 0 = B + D 0 = A + 5 B + C - 5 D 0 = 10 A - 25 B - 10 C - 25 D 1 = 25 A - 125 B + 25 C + 125 D

Swap the sides

⎩ ⎨ ⎧ B + D = 0 A + 5 B + C - 5 D = 0 10 A - 25 B - 10 C - 25 D = 0 25 A - 125 B + 25 C + 125 D = 1

Solve the equation for B

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⎩ ⎨ ⎧ B = - D A + 5 B + C - 5 D = 0 10 A - 25 B - 10 C - 25 D = 0 25 A - 125 B + 25 C + 125 D = 1

Substitute the given value of B into the equation ⎩ ⎨ ⎧ A + 5 B + C - 5 D = 0 10 A - 25 B - 10 C - 25 D = 0 25 A - 125 B + 25 C + 125 D = 1

⎩ ⎨ ⎧ A + 5 (- D) + C - 5 D = 0 10 A - 25 (- D) - 10 C - 25 D = 0 25 A - 125 (- D) + 25 C + 125 D = 1

Simplify

⎩ ⎨ ⎧ A - 10 D + C = 0 10 A - 25 (- D) - 10 C - 25 D = 0 25 A - 125 (- D) + 25 C + 125 D = 1

Simplify

⎩ ⎨ ⎧ A - 10 D + C = 0 10 A - 10 C = 0 25 A - 125 (- D) + 25 C + 125 D = 1

Simplify

⎩ ⎨ ⎧ A - 10 D + C = 0 10 A - 10 C = 0 25 A + 250 D + 25 C = 1

Solve the equation for A

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⎩ ⎨ ⎧ A = 10 D - C 10 A - 10 C = 0 25 A + 250 D + 25 C = 1

Substitute the given value of A into the equation {10 A - 10 C = 0 25 A + 250 D + 25 C = 1

{10 (10 D - C) - 10 C = 0 25 (10 D - C) + 250 D + 25 C = 1

Simplify

{100 D - 20 C = 0 25 (10 D - C) + 250 D + 25 C = 1

Simplify

{100 D - 20 C = 0 500 D = 1

Solve the equation for D

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{100 D - 20 C = 0 D = \frac{1}{500}

Substitute the given value of D into the equation 100 D - 20 C = 0

100 \times \frac{1}{500} - 20 C = 0

Multiply the numbers

\frac{1}{5} - 20 C = 0

Move the constant to the right-hand side and change its sign

- 20 C = 0 - \frac{1}{5}

Removing 0 doesn't change the value,so remove it from the expression

- 20 C = - \frac{1}{5}

Change the signs on both sides of the equation

20 C = \frac{1}{5}

Multiply by the reciprocal

20 C \times \frac{1}{20} = \frac{1}{5} \times \frac{1}{20}

Multiply

C = \frac{1}{5} \times \frac{1}{20}

Multiply

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C = \frac{1}{100}

Substitute the given values of D, C into the equation A = 10 D - C

A = 10 \times \frac{1}{500} - \frac{1}{100}

Simplify the expression

A = 10 \times 50 0^{- 1} - 10 0^{- 1}

Calculate

A = \frac{1}{100}

Substitute the given value of D into the equation B = - D

B = - \frac{1}{500}

Calculate

⎩ ⎨ ⎧ A = \frac{1}{100} B = - \frac{1}{500} C = \frac{1}{100} D = \frac{1}{500}

Substitute back

\frac{1}{100 ( t - 5 ) ^{2}} - \frac{1}{500 t - 2500} + \frac{1}{100 ( t + 5 ) ^{2}} + \frac{1}{500 t + 2500}

5 \times \int (\frac{1}{100 ( t - 5 ) ^{2}} - \frac{1}{500 t - 2500} + \frac{1}{100 ( t + 5 ) ^{2}} + \frac{1}{500 t + 2500}) d t

Use the property of integral \int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

5 (\int \frac{1}{100 ( t - 5 ) ^{2}} d t + \int - \frac{1}{500 t - 2500} d t + \int \frac{1}{100 ( t + 5 ) ^{2}} d t + \int \frac{1}{500 t + 2500} d t)

Calculate

5 \times \int \frac{1}{100 ( t - 5 ) ^{2}} d t + 5 \times \int - \frac{1}{500 t - 2500} d t + 5 \times \int \frac{1}{100 ( t + 5 ) ^{2}} d t + 5 \times \int \frac{1}{500 t + 2500} d t

Evaluate the integral

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- \frac{1}{20 t - 100} + 5 \times \int - \frac{1}{500 t - 2500} d t + 5 \times \int \frac{1}{100 ( t + 5 ) ^{2}} d t + 5 \times \int \frac{1}{500 t + 2500} d t

Evaluate the integral

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- \frac{1}{20 t - 100} - \frac{1}{100} ln (∣ t - 5 ∣) + 5 \times \int \frac{1}{100 ( t + 5 ) ^{2}} d t + 5 \times \int \frac{1}{500 t + 2500} d t

Evaluate the integral

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- \frac{1}{20 t - 100} - \frac{1}{100} ln (∣ t - 5 ∣) - \frac{1}{20 t + 100} + 5 \times \int \frac{1}{500 t + 2500} d t

Evaluate the integral

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- \frac{1}{20 t - 100} - \frac{1}{100} ln (∣ t - 5 ∣) - \frac{1}{20 t + 100} + \frac{1}{100} ln (∣ t + 5 ∣)

Solution

- \frac{1}{20 t - 100} - \frac{1}{100} ln (∣ t - 5 ∣) - \frac{1}{20 t + 100} + \frac{1}{100} ln (∣ t + 5 ∣) + C, C \in R