Monday

September 22, 2014

September 22, 2014

Posted by **Vivian** on Wednesday, February 8, 2012 at 1:35am.

- Calculus -
**Reiny**, Wednesday, February 8, 2012 at 8:23amI split (x+3)/(x^2+9) into

x/(x^2 + 9) + 3/(x^2+9)

= (1/2)(2x/(x^2+9) + 3/(x^2 + 9)

the integral of the first is (1/2) ln(x^2+9)

the second I looked up in my old integral tables and I would get

tan^2 (x/3)

so ∫[(x+3)/(x^2 +9)] dx

= (1/2)ln(x^2+9) + tan^1 (x/3)

**Answer this Question**

**Related Questions**

Calculus 2 correction - I just wanted to see if my answer if correct the ...

Calculus - Evaluate the triple integral ∫∫∫_E (x+y)dV where E ...

Calculus - Evaluate the triple integral ∫∫∫_E (x)dV where E is...

Calculus - Find the integral by substitution ∫ [(16 x3)/(x4 + 5)] dx &#...

Calculus - Evaluate the triple integral ∫∫∫_E (xy)dV where E ...

Calculus - Evaluate the triple integral ∫∫∫_E (z)dV where E is...

Calculus - Evaluate the triple integral ∫∫∫_E (xyz)dV where E ...

Calculus - I'm having trouble reversing the order of integration of ∫&#...

calculus (check my work please) - Not sure if it is right, I have check with the...

Help Evaluating Integrals - 1.) ∫ (2)/(x-4) dx 2.) ∫ sec^2x tanx dx ...