Question

Hello. Frustrating doing the calculus because the text always jumps yo a conclusion without adequately explaining the preliminary steps to solve the equation.

∫ (8x + 1) (2x^2 - x - 1)^-1

When I factor out the integral I get:

3 ∫ (8x + 1)^-1 + 2 ∫ (x- 1)^-1


Whereas the text and wolframalpha place the constants factored outside the integrals reversed.

2 in front of (2x + 1) and 3 in front of (x - 1)


I suppose there is a simple answer. Anyway explaining the problem as a ? Helps me understand it



whereas the book has the constants outside the integrals reversed

Answer 1

\(\dfrac{8x+1}{(2x+1)(1-x)}= \dfrac{A}{2x+1}+\dfrac{B}{1-x}\)
what you got A and B values as ? which is 2 and which is 3 ?

Answer 2

to find A, put x = -1/2
to find B, put x=1

Answer 3

in \(8x+1=A(1-x)+B(2x+1)\)

Answer 4

3 ( 2x + 1) and 2 ( x - q)
Or 3 times the 2x gets 6 x and 2 times the x gets 2x which totals 8 x

Answer 5

\(8x+1=A(1−x)+B(2x+1)\)
x=1 gives , B =3
x=-1/2 gives A =-2
so,

\(8x+1=-2(1−x)+3(2x+1)\)
or
\(8x+1=2(x-1)+3(2x+1)\)
yes, you are correct, so whats the doubt ?

Answer 6

\(\dfrac{8x+1}{(2x+1)(1-x)}= \dfrac{-2}{2x+1}+\dfrac{3}{1-x}\)

Answer 7

Book says the opposite. 2 ( 2x + 1) and 3 ( x- 1)

Answer 8

that must be the final answer ?

Answer 9

Thanks for the help. I 'll look into my error.

Answer 10

ask if you still didn't get it...

Answer 11

i see now that i have taken 1-x instead of x-1 :P
then your A will be 2
and
\(\int \dfrac{8x+1}{(2x+1)(x-1)}= \int \dfrac{2}{2x+1}+\dfrac{3}{x-1} \\ =2/2 \ln (2x+1)+3 \ln (x-1)+c\)

Answer 12

My error was not using the neg 1 exponent to make it a fraction. Thanks sometimes defining the ? Is as important as discovering the answer.

Answer 13

ohh..ok, welcome :)