Q1)? x / (x-1)(x^2 + 4)dxQ 2)? x^2 + 2x + 3 / whole root over x^2 + 1Q 3)?e^ax cos bx dxQ 4)?dx / 1+3e^x + 2e^2xdxQ 5)?x^3 + 1 / x^3 – 1dxQ 6)?x^2 / x^4 – x^2 – 12 dx

(1) x / (x-1)(x^2 + 4)dx

integral  [1/5.(x-1) -  2x/10(x²+4)  + 4/5(x²+4)] .dx

solution is = (ln(x-1)) /5 - (ln(x²+4)) /10 +2/5.tan^-1(x/2) +C where C is constant

(2) x^2 + 2x + 3 / whole root over x^2 + 1

integral  [ (x²+1-1)/(x²+1)^1/2 +  2x/(x²+1)^1/2  + 3/((x²+4)^1/2)] .dx

integral  [ (x²+1)^1/2 +  2x/(x²+1)^1/2  + 2/((x²+4)^1/2)] .dx

solution. is = [x.(x²+1)^1/2]/2 + 2.(x²+1)^1/2 + 5/2*ln(x+(x²+1)^1/2 ) +C where C is constant

(3) (e^ax cos bx) dx

integral P = e^ax.cos(bx). dx

integrate by parts with M=e^ax     dm = ae^ax.dx

P =1/b*e^ax.sin(bx) - a/b* [ integral of e^ax.sin(bx). dx ]

again integrate by parts gives

integral of  e^ax.sin(bx). dx = -1/b*e^ax.cos(bx) + a/b*P

Putting into above expression gives

P = 1/b*e^ax.sin(bx) +2/b²*e^ax.cos(bx) - (a/b)²*P

P = b² / (a²+b²) .e^ax.(sin bx /b + acos(bx) /b² ) + C

P = e^ax / (a²+b²)^1/2.(cos(bx) - tan^-1(b/a) ) + C where C is constant

(4) dx / (1+3e^x + 2e^2x)  .dx

integral  [1/(1+3e^x +2e^2x)] .dx

[1/(1+2e^2x)*(e^x+1)] .dx

2.dx/(1+2e^2x)  - dx / (e^x+1)

put  1+2e^2x =t and e^x+1 = k

dt / (t-1)  - dt /t + dk / k - dk/(k-1)

ln(2e^x) - ln(1+2e^x) +ln(1+e^x) - ln(e^x) +C

final solution is :  ln(2.(1+2e^x)/(1+e^x)) +C where C is constant

(5) (x^3 + 1) / (x^3 – 1) .dx

also can be written as : 1 +( 2/(x³-1)) .dx

we know that : x³-1 = (x-1)(x²+x+1)

let P = integral of   (1/(x³-1)) .dx

hence, by partial fractions:

1/(x³-1) = A/(x-1) + (Bx+C)/(x²+x+1); hence,
1 = A(x²+x+1) + (Bx+C)(x-1) =
(Ax² + Ax + A) + (Bx² + (C-B)x - C) =
(A+B)x² + (A+C-B)x + (A-C); hence,
A + B = 0, A + C - B = 0, and A - C = 1; hence,
A = - B, C = B - A = -2A, and A -(- 2A) = 1; hence,
A = 1/3, B = -1/3, and C = -2/3

Hence,

P= ∫(1/(x³-1))dx = ∫[(1/3)/(x-1)] + [((-1/3)x-2/3)/(x²+x+1)]dx
   = ∫[(1/3)/(x-1)]dx + ∫[((-1/3)x-2/3)/(x²+x+1)]dx
   = (1/3)∫[1/(x-1)]dx + (-1/3)∫[(x+2)/(x²+x+1)]dx
   = (1/3)∫[1/(x-1)]d(x-1) + (-1/3)∫[(x+(1/2)+(3/2))/((x+(1/2))²+(√3/2)²) dx
   = [(1/3)ln|x-1|  - (1/3)∫[(x+(1/2)+(3/2))/((x+(1/2))²+(√3/2²) + C
   = [(1/3)ln|x-1|  - (1/3)∫[(x+½)/((x+½)²+(√3/2)²]dx - (1/3)(3/2)∫[1/((x+½)²+(√3/2)²)]dx
   = (1/3)ln|x-1|  - (1/6)ln|(x+½)²+(√3/2)²| - (½)[(1/(√3/2))arctan((x+½)/(√3/2))]  + C
   = (1/3)ln|x-1|  - (1/6)ln((x+½)²+(√3/2)²) - (1/√3)arctan((2x+1)/√3) + C

so final solution is : =  x + (2/3)ln|x-1|  - (1/3)ln((x+½)²+(√3/2)²) - (2/√3)arctan((2x+1)/√3) + C where C is constant

(6) x^2 / x^4 – x^2 – 12 dx

= 1/7* [ 7x² / (x²-4)(x²+3)] .dx

= 4/7* [ 1 / (x²-4)] + 3/7*[1 / (x²+3)] .dx

solution. is = 1/7*ln[(x+2)/(x-2)] + 3^1/2/7 * tan^-1(x/3) +C where C is constant