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# Calculus Help?

Ive been having some trouble solving these problems. Any help would be absolutely appreciated.

I figured out #3.

Figured out 1 & 3.

### 1 Answer

- 2 months ago
1) The problem is Int(x*sin2x)dx

Integrate by parts

Int(u *dv) = uv - int(v * du)

We need to indentify the parts now.

u = x

dv = sin2x

v = int(sin2x)dx = -1/2 * cos(2x)

du =1

Now plug it all into the equation Int(u *dv) = uv - int(v * du)

Int(x*sin2x)dx = -1/2*x*cos(2x) - int(-1/2 *cos(2x))

= -1/2*x*cos(2x) + 1/4 sin(2x)

=1/4(sin(2x) - 2x*cos(2x)) + c

I think I did this right

2) I'm not really sure how to do this, it's been too long

3) Int((x-6)/(x(x-3)) dx

First, simplify the integrand using partial fraction decomposition, and you end up with

Int(2/x - 1/(x-3))dx

= Int(2/x) dx - Int((x-3)^-1)dx

= 2 Int x^-1 dx - int ((x-3)^-1)dx

= 2 ln x - ln (x-3) + c

4) Not quite sure

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