Express the integrated as a sum of partial fractions and evaluate the integral
3x2+x+9/(x2+5)(x-6)
The partial fractions part come to:
(3x2+x+9)/((x2+5)(x-6))
= 3/(x-6) + 1/(x^2+5)
both of those are easy to integrate, right?
We must be careful to insert parentheses when they are needed.
Expressions are evaluated according to the PEMDAS rule.
http://www.purplemath.com/modules/orderops.htm
For example, what you posted is mathematically equivalent
3x²+x +[9/(x²+5)]*(x-6)
which is (probably) not what you meant.
Correctly written, what you (probably) meant should read:
(3x2+x+9)/((x2+5)(x-6))
Here, we try to interpret correctly by guessing from context, but calculators are unforgiving with missing parentheses, that is why we need to develop the proper reflex for them.
To express the integral as a sum of partial fractions, we need to decompose the given rational function into simpler fractions.
First, let's factorize the denominator, x^2 + 5, which cannot be factored further.
Next, we factorize the second term, (x - 6), which is already in its factored form.
Now, we can express the given rational function as a sum of partial fractions:
3x^2 + x + 9 / [(x^2 + 5)(x - 6)] = A / (x^2 + 5) + B / (x - 6)
To find the values of A and B, we need to clear the denominators:
3x^2 + x + 9 = A(x - 6) + B(x^2 + 5)
Now, let's solve for A and B.
Comparing the coefficients of x^2, we get:
0 = B
Comparing the coefficients of x^1, we get:
1 = -6A
Solving for A:
A = -1/6
Therefore, the partial fraction decomposition becomes:
3x^2 + x + 9 / [(x^2 + 5)(x - 6)] = -1/6 / (x^2 + 5) + B / (x - 6)
Now, we can integrate the partial fractions separately:
∫ [-1/6 / (x^2 + 5)] dx + ∫ [B / (x - 6)] dx
The first integral can be evaluated using the inverse tangent function, and the second integral is a simple logarithmic function.
To evaluate the integral, we need the limits of integration. Please provide the limits, and I can provide the final result.