0.
Well I'd be interested if you solve some of these, if you can solve the first one int (e
cos(x)^2 *In(x) * cos(x) dx) I'd be very, very impressed!
I can solve a few, here's one I'm pretty sure I can solve
int((7x + 10) / (x - 5)(x
2 + 2x + 10) dx
First, parfrac it (because we have an irreducable quadratic we'll have to stick with it)
(7x + 10) / (x - 5)(x
2 + 2x + 10) = A / (x - 5) + (Bx + C) / (x
2 + 2x + 10)
Multiply both sides by the denominator on LHS
7x + 10 = A(x
2 + 2x + 10) + (Bx + C)(x - 5)
CASES: First make x = 5
7(5) + 10 = A(25 + 2(5) + 10) + 0
45 = 45A
A = 1
Now, use x = 0
10 = (1)(10) + C(-5)
0 = -5C
C = 0
Last case can be anything, lets just say x = 1
17 = (1)(13) + (

(-4)
17 - 13= -4B
4 = -4B
B = -1
14/7 = 4B
B = 14/28
B = 1/2
Therefore the parfrac is 1/(x- 5) - x/(x
2 + 2x + 10)
Now we can INTEGRATE each individual parfrac
int (1/x-5)dx is ln(x - 5)
int (-x/x
2 + 2x + 10) is -1/2(ln(x
2 + 2x + 10) but in order to get rid of that pesky 2 in the numerator of the integral we can use the inverse tangent equations, to get (1/3)*tan
-1((1/3)*x + 1/3)
So our final integral is
ln(x - 5) - 1/2*ln(x
2 + 2x + 10) + (1/3)*tan
-1((1/3)*x + 1/3) + C
I only know that because we did Partial fractions last semester and it's one of the things I remember xD