Determine the following intergrals:

Question

1. ∫ 3x/(x+1)³ dx (by using substitution)

2. ∫ (4x² - 4x - 3)/ [x²(x-3)] dx (by using partial intergration)

3. ∫ e^2xcos(2x) dx (by using integration by parts)

Answer 1

a) Int 3xdx/(x+1)³ _{using substitution let u=x+1 then du =dx x+1=u x= u - 1 integral becomes 3 (u-1) du/u}³ this gives

Integral of 3du/u² - 3du/u³ which equals -3/u +3/2 u² which becomes 3/2(x+1)² -3/(x+1) +C this gives

(-6x - 3)/2(x+1)² +C

b) Int(4x² -4x -3)/x²(x-3) by partial fractions use A/x +B/x² +C/(x-3) this becomes Ax² -3Ax +Bx -3B +Cx²

matching coefficients gives 4=A+C, -4=-3A +B, -3B= -3. Then B=1 A=5/5 and C=7/3 So the integral becomes Int( 5dx/3x +dx/x² +7dx/3(x-3) ) Doing the integrals gives 5log|x|/3 -1/x +7log|x-3|/3 + C

c) Int e^2xcos(2x)dx by parts let u=e^2x dv= cos(2x) then du = 2e^2x dx and v= sin(2x)/2 so Int = e^2xsin(2x)/2 - Int e^2x sin(2)dx let u=e^2x and dv= sin(2x)dx so again du =2e^2xdx and v= - cos(2x)/2 so now the original integral equals e^2x sin(2x)/2 + e^2xcos(2x)/2 - Int e^2xcos(2x)dx moving this last term to the other side gives

2Int e^2xcos(2x)dx = e^2xsin(2x)/2 +e^2xcos(2x)/2 +C. Therefore Int e^2xcos(2x)dx = e^2xsin(2x)/4 + e^2xcos(2x)/4 + C.

Answer 2

In the first one u = x+1 will get you 2 integrals will simple powers in the denominator.

The second one will be done by using partial fractions of the form (Ax + B)/x² + C/(x-3).

You will get 3 linear equations in the variables A, B and C and the term in Ax + B will require a substitution to get a single power in the denominator.

In the 3rd one make the substitution u=2x first, then the integration by parts should be easy.

If you are still stuck, let me know and I will try to help further.