Struggling with integration by parts and partial fractions! Plz help!?

Question

Use Intergration by parts to

Pie

Integrate 4x sin 3x dx

0

Use the method of partial fractions to show that

10x-2/(x-5)(x+3) = 6/x-5+4/x+3

Hence find the area under the curve y= 10x-2/(x-5)(x+3)

Between x=6 and x=9 and the x-axis.

Answer 1

Let S be the symbol of integral.

S 4x sin 3x dx

dv = sin 3x dx

v = -1/3 cos 3x

S 4x sin 3x dx = 4x . (-1/3 cos 3x) - S (-1/3 cos 3x) . 4 dx

=-4/3 cos 3x + 4/3 S cos 3x dx

=-4/3 cos 3x + 4/3 . 1/3 sin 3x + c

Put in the numbers...

(-4/3 cos 3.pi + 4/9 sin 3.pi) - (-4/3 cos 0 + 4/9 sin 0) = 8/3

(10x-2)/(x-5)(x+3) = A/(x-5) + B/(x+3) = (A(x+3) + B(x-5))/(x-5)(x+3)

10x-2 = Ax+3A+Bx-5B = (A+B)x + (3A-5B)

A+B = 10

3A-5B = -2

----------------

3A+3B = 30

3A - 5B = -2

----------------

8B = 32

B = 4

A = 6

so,

(10x-2)/(x-5)(x+3) = A/(x-5) + B/(x+3) = 6/(x-5) + 4/(x+3)

For the area, its the same as integral from 6 to 9 y dx

S 10x-2/(x-5)(x+3 )dx = S  6/x-5+4/x+3 dx = 6.ln|x-5| + 4.ln|x+3| + c

Put in the number...

(6.ln 3 + 4.ln 9) - (6.ln 4 + 4.ln 12) = 6.ln(3/4) - 4.ln(9/12) = 2.ln(3/4)

Area = 2.ln(3/4)

Answer 2

Too log man cannt be typed here  

Answer 3

Aldrian says right

blindly follow it