The integral of [(x+3)/(x^2+9)] dx
I think you split it into two fractions but im not exactly sure what to do.
yes, you can split it into
x/(x^2 + 9) + 3/(x^2+9)
the first part becomes
(1/2) ln(x^2 + 9)
and the second part becomes
sin^-1 (3/x) .... (had to look that one up in my old-fashioned table of integrals)
don't forget to add the constant
Hmmm. I get tan^-1 (x/3)
Steve, thanks for the check-up,
you are right, I must have had a senior moment.
To integrate the given expression, you can indeed split it into two fractions. Let's break down the process step by step:
1. Begin by factoring the denominator of the fraction. In this case, the denominator is x^2 + 9, which cannot be factored further.
2. Write the fraction as the sum of two fractions with distinct numerators. This can be done by expressing the numerator, x + 3, as the sum of two parts.
x + 3 = x + 0 + 3
3. Rewrite the initial fraction by splitting the numerator into two parts:
[(x + 0) / (x^2 + 9)] + [(3) / (x^2 + 9)]
4. Each of these fractions can be integrated separately as its own term.
∫[(x + 0) / (x^2 + 9)] dx + ∫[(3) / (x^2 + 9)] dx
5. Evaluate the integral of each term individually.
For the first term ∫[(x + 0) / (x^2 + 9)] dx:
Let u = x^2 + 9.
Taking the derivative of both sides, du = 2x dx, which gives us x dx = (1/2) du.
Substituting for x dx in the integral, we have (1/2) ∫[(1/u)] du.
Integrate the term using the inverse of the natural logarithm: (1/2) ln|u| + C.
Since we initially let u = x^2 + 9, substitute back for u: (1/2) ln|x^2 + 9| + C1.
For the second term ∫[3 / (x^2 + 9)] dx:
Since the integral of 1/(x^2 + a^2) is arctan(x/a), let a = 3.
Integrate the term using the arctangent function: 3 * arctan(x/3) + C.
6. Combine the solutions for each term.
The integral of [(x+3)/(x^2+9)] dx is equal to (1/2) ln|x^2 + 9| + 3 * arctan(x/3) + C2, where C2 is the constant of integration.
Remember to always include the constant of integration when finding indefinite integrals.