Question

Please help!
Evaluate the integral

Answer 1

\[\int\limits_{}^{}\frac{ x+1 }{ (2-x)(x^{2}+4) }\] with respect to x

Answer 2

partial fractions

Answer 3

Would the partial fractions be\[\frac{ A }{ 2-x }+\frac{ Bx+C }{ x^2+4 }\]

Answer 4

yes

Answer 5

continue @roberts.spurs19

Answer 6

so then A(x^2) + (Bx+c)(2-x) = x+1
And then i compare coefficients

Answer 7

This gives me A=B=3/8 and C=-1/4

Answer 8

hmmm It 's been a while

no its
A(x^2 + 4) + (Bx+c)(2-x) = x+1

Answer 9

yep its (x^2+4)

Answer 10

Sorry I did mean that!

Answer 11

so continue
find the values for A and B

Answer 12

yea those values of A, B and C are what i get

Answer 13

Ok so then i integrate \[\int\limits_{}^{}\frac{ 3 }{ 8(2-x } +\int\limits_{}^{}\frac{\frac{ 3 }{ 8 }x+\frac{ 1 }{ 4 } }{ x^2 +4 }\]

Answer 14

that should be - 1/4

Answer 15

Sorry thanks!
Then this gives me \[\frac{ 3 }{ 16 }\ln x + \frac{ 1 }{ 8 } \int\limits\limits_{}^{\frac{ }{ }}\frac{ 3x -2 }{ x^2 +4 }\]

Answer 16

should i do partial fractions again or is there an easier way please?

Answer 17

Sorry i just realised that you can't !

Answer 18

\[1/8 \int\limits_{}^{}\frac{ 3x }{ x^2 +4 } +\frac{ 2 }{ x^2+4 }\]

Answer 19

no - you are right
I'd have to look this up
the integral of 1 / (x^2 + 1) is tan-1 so that will come into somewhere

Answer 20

ok so the second part goes to \[\frac{ 1 }{ 8 } \tan^{-1} \frac{ x }{ 2}\]

Answer 21

yes looks good

Answer 22

I wonder can you convert the first part to f'(x) / f(x) ?

Answer 23

so all thats left is \[\frac{ 1 }{ 8 }\int\limits_{}^{}\frac{ 3x }{ x^2 +4 }\]

Answer 24

convert to