Question

integration by partial fractions doubt

Answer 1

how do we handle the foll to solve for A,B and C?

\[\Large 2x^2+3=\frac{A}{x-1}\;\;\;+\frac{B}{x+2}\;\;+\frac{C}{x^2+4}\]

Answer 2

@Hero , @dpaInc

Answer 3

Didn't the left side have a denominator?

Answer 4

ya soory

Answer 5

u can guess what it was lols

Answer 6

Well it would sure be helpful if you would post the original problem accurately

Answer 7

\[\Large \int\limits \frac{2x^2+3\;\;dx}{(x^2-1)(x^2+4)}\]

Answer 8

\[\frac{2x^2+3}{((x-1)(x+2)(x^2+4)}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{Cx+D}{x^2+4}\]

Answer 9

Try that.

Answer 10

i dont knw hw to do this

Answer 11

Multiply both sides by the common denominator.

Answer 12

why did u write Cx+d?

Answer 13

\[2x^2+3=A(x+1)(x^2+4)+B(x-1)(x^2+4)+(Cx+D)(x-1)(x+1)\]

Answer 14

Because the denominator is an irreducilbe quadratic factor

Answer 15

@mertsj i didnt get tht can u explain?

Answer 16

Do you see that one factor of the denominator is x^2+4?

Answer 17

It cannot be factored any further and it is quadratic so the partial fraction decomposition will have a term of the form:
\[\frac{Ax+B}{ax^2+bx+c}\]

Answer 18

only one term?

Answer 19

What do you mean "only one term"?

Answer 20

i mean here we took A and B as numbers and C as a linear fn hw did we knw only one should be linear?