Saturday

January 31, 2015

January 31, 2015

Posted by **Hannah** on Saturday, April 30, 2011 at 6:06pm.

integral of 3 to 2 x/(x^2-2)^2 dx

u=x^2-2

du=2x dx

1/2 du = x dx

integral of 1/u^2 du

-1/(x^2-2) Then I plug in 3 and 2 and subtract them form each other

-1/(3^2-2) - (-1/(2^2-2) Is this correct?

- Math(Please check) -
**MathMate**, Saturday, April 30, 2011 at 9:09pmThe (1/2) has been missed out, should read:

∫ du/(**2**u^2)

=-1/(**2**(x^2-2))

After that you can plug in the limits of integration. I get 5/28 integrating from 2 to 3.

The bottom limit is the start value, and the top value is the end value.

It is usual to go from 2 to 3, and rarely from 3 to 2. Please check.

**Answer this Question**

**Related Questions**

calc asap! - can you help me get started on this integral by parts? 4 S sqrt(t) ...

Calculus II/III - A. Find the integral of the following function. Integral of (x...

Calculus - evaluate the integral or state that it diverges. Check if I did it ...

calc check - <<y=(1/A)*integral from a to b of: (1/2)[f(x)]^2 dx >> ...

Calculus - integral -oo, oo [(2x)/(x^2+1)^2] dx (a) state why the integral is ...

calculus - 8). Part 1 of 2: In the solid the base is a circle x^2+y^2=16 and the...

math - evaluate the double integral and reverse order of integration [(first ...

calc - how do you start this problem: integral of xe^(-2x) There are two ways: 1...

math - Evaluate the following indefinite integral by using the given ...

math - Evaluate the following indefinite integral by using the given ...