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```        Q1)   		? x / (x-1)(x^2 + 4)dx
Q 2)           ? x^2 + 2x + 3 / whole root over x^2 + 1

Q 3)    ?  e^ax cos bx dx

Q 4)   ?  dx / 1+3e^x + 2e^2xdx

Q 5)  ?    x^3 + 1 / x^3 – 1dx

Q 6)    ?     x^2 / x^4 – x^2 – 12 dx
```
10 years ago

```							(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 = eax.cos(bx). dx
integrate by parts with M=eax     dm = aeax.dx
P =1/b*eax.sin(bx) - a/b* [ integral of eax.sin(bx). dx ]
again integrate by parts gives
integral of  eax.sin(bx). dx = -1/b*eax.cos(bx) + a/b*P
Putting into above expression gives
P = 1/b*eax.sin(bx) +2/b2*eax.cos(bx) - (a/b)2*P
P = b2 / (a2+b2) .eax.(sin bx /b + acos(bx) /b2 ) + C
P = eax / (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+3ex +2e2x)] .dx
[1/(1+2e2x)*(ex+1)] .dx
2.dx/(1+2e2x)  - dx / (ex+1)
put  1+2e2x =t and ex+1 = k
dt / (t-1)  - dt /t + dk / k - dk/(k-1)
ln(2ex) - ln(1+2ex) +ln(1+ex) - ln(ex) +C
final solution is :  ln(2.(1+2ex)/(1+ex)) +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)] + 31/2/7 * tan-1(x/3) +C where C is constant

```
10 years ago
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