Integration problem?

Question

How to integrate x^2/(x^2-4) dx and 1/(x^3+x) dx?Give answer with explaination.

Answer 1

The hint from the others are very good.
But here I will show you an algebraic form to solve these kind of integrals:

a)
∫x² /(x² -4) dx
We can express x² /(x² -4) like
(x² -4+4) / (x² -4)
=1 + 4 / (x² -4)..........(I)

But 4 / (x² -4) = 4 /[ (x-2)(x+2)]
and we can write in the form of

4 / [(x -2)(x+2)] = A /(x-2) + B /(x+2)
Solving
4= A(x+2) + B(x-2);
4 = x(A+B) +2(A-B) →A+B =0 -because there is no x-
and 2(A-B) = 4
Being A = -B → 2(2A) = 4 and A = 1 and B= -1
Then using in (I)
=1 + 4 / (x² -4) = 1+ 1 /(x-2) -1 /(x+2)

Integrating:
∫{1 + 4 / (x² -4)} dx= ∫{1+ 1 /(x-2) -1 /(x+2) }dx
=∫dx+ ∫1/ (x-2) dx -∫1 /(x+2) dx
= x + ln[x-2] - ln[x+2] +C
= x + ln[ (x-2) /(x+2)] + C

b) ∫1/(x³+x) dx = ∫1 / [x(x²+1)] dx
Again using
1/[x(x²+1)] = A/x + (Bx+C) / (x²+1)

→ 1 = A(x²+1) +(Bx+C) x = (A+B)x² + Cx + A = 1
But
A+B=0 → A= -B
A = 1
B = -1
C = 0 then

1/[x(x²+1)] = 1/x - x/(x²+1)

∫1/(x³+x) dx = ∫{1/x - x/(x²+1) }dx

= ln[x] - 1/2ln[x²+1] +C

= ln [x / (x²+1)] +C

Derive these results and you will see that are correct.

Answer 2

For the first one,

int (x^2/x^2 - 2) dx
= int (1 + 4/(x^2 - 4)) dx
= int (1 + 4 (1/(x-2) - 1/(x + 2)))) dx
= x + 4ln(x-2) - 4 ln(x+2) + C
= x + 4 ln ((x-2) / (x+2)) + C

For the second one,

int (1/(x^3 + x)) dx
= int ( x / (x^2 (x^2+1)) ) dx
= 1/2 int ( 1 / (x^2 (x^2+1)) ) dx^2
= 1/2 int ( 1 / (y (y+1)) ) dy (Let y = x^2)
= 1/2 int (1/y - 1/(y+1) ) dy
= 1/2 ln(y) - ln(y+1) + C
= 1/2 lnx^2 - 1/2 ln(x^2 + 1) + C
= lnx - 1/2 ln(x^2 + 1) + C

Answer 3

1. Add & subtract 4 in numerator

2. Club -4 with x^2 & separate it with +4

3. Now , first integral will give u x & second one 4/(x^2-4)

4. now second one can be written as [ {1/(x-2)} - {1/(x+2)}]

5. now the result can be obtained by using log formulae .


Tell me if u got it now

BYE

Answer 4

cause