# How do you find #int 10/((x-1)(x^2+9)) dx# using partial fractions?

Since the denominator consists of 1 linear factor and 1 irreducible quadratic factor, we may use the method of partial fractions to rewrite it as :

Now comparing terms we get that

Solving this linear system of equations yields :

Therefore the original integral may be written and solved as

By signing up, you agree to our Terms of Service and Privacy Policy

To find the integral of ( \frac{10}{(x-1)(x^2+9)} ) using partial fractions, we first decompose the fraction into partial fractions. We express ( \frac{10}{(x-1)(x^2+9)} ) as ( \frac{A}{x-1} + \frac{Bx + C}{x^2+9} ). Then, we find the values of A, B, and C by equating numerators.

Multiplying both sides by the denominator ( (x-1)(x^2+9) ), we get: [ 10 = A(x^2 + 9) + (Bx + C)(x - 1) ]

Expanding and collecting like terms, we get: [ 10 = Ax^2 + 9A + Bx^2 - Bx + Cx - C ]

Matching coefficients for like terms, we have: [ (A + B)x^2 - Bx + Cx + (9A - C) = 10 ]

This equation implies that: [ A + B = 0 ] [ -B + C = 0 ] [ 9A - C = 10 ]

Solving this system of equations, we find: [ A = \frac{5}{3} ] [ B = -\frac{5}{3} ] [ C = -\frac{5}{3} ]

Now, we rewrite the original integral using the partial fraction decomposition: [ \int \frac{10}{(x-1)(x^2+9)} ,dx = \int \left( \frac{5}{x-1} - \frac{5x + 5}{x^2 + 9} \right) ,dx ]

We can now integrate each term separately: [ \int \frac{5}{x-1} ,dx = 5\ln|x-1| + C_1 ] [ \int \frac{-5x - 5}{x^2 + 9} ,dx = -\frac{5}{2}\ln|x^2 + 9| + C_2 ]

Where ( C_1 ) and ( C_2 ) are constants of integration.

Therefore, the integral of ( \frac{10}{(x-1)(x^2+9)} ) using partial fractions is: [ 5\ln|x-1| - \frac{5}{2}\ln|x^2 + 9| + C ]

By signing up, you agree to our Terms of Service and Privacy Policy

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7