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Question

(8x^2 - 5x + 1/sq rt 8) / (x-2) (x^2 + 4)

i've tried substitution but got stuck. any help would be great. thank you.

Answer 1

(8x^2 - 5x + 1/√8) / [(x-2) (x^2 + 4)]
Use partial fractions to write this as A / (x-2) + (Bx + C) / (x^2 + 4)
= [A(x^2 + 4) + (Bx + C)(x-2)] / [(x-2) (x^2 + 4)]
So we have
Ax^2 + 4A + Bx^2 + Cx - 2Bx - 2C = 8x^2 - 5x + 1/√8
<=> (A+B)x^2 + (C-2B)x + (4A-2C) = 8x^2 - 5x + 1/√8
<=> A+B = 8, C-2B = -5, 4A-2C = 1/√8.
From the first two equations we get 2A+C = 11, so 4A+2C = 22. Combining this with the last gives 8A = 22 + 1/√8, so A = 11/4 + 1/(16√2), and 4C = 22 - 1/√8, so C = 11/2 - 1/(8√2). Then B = 8 - A = 21/4 - 1/(16√2).

So we have
∫(8x^2 - 5x + 1/√8) / [(x-2) (x^2 + 4)] dx
= ∫(11/4 + 1/(16√2)) / (x-2) dx + [(21/4 - 1/(16√2))x + (11/2 - 1/(8√2))] / (x^2 + 4) dx
= (11/4 + 1/(16√2)) ln |x-2| + (21/8 - 1/(32√2)) ∫ (2x / (x^2+4)) dx + [11/2 - 1/(8√2)] ∫ (1 / (x^2+4)) dx
= (11/4 + 1/(16√2)) ln |x-2| + (21/8 - 1/(32√2)) ln (x^2+4) + [11/2 - 1/(8√2)] (1/2) arctan (x/2) + c
= (11/4 + 1/(16√2)) ln |x-2| + (21/8 - 1/(32√2)) ln (x^2+4) + [11/4 - 1/(16√2)] arctan (x/2) + c.

Pretty ugly, but with that 1/√8 sitting there it was never going to be anything else.

Answer 2

The integral is:

-(1/64)*ln(x^2+4)*2^(1/2)+(21/8)*ln(x^2+4)+(11/4)*arctan(1/2*x)
-(1/32)*arctan(1/2*x)*2^(1/2)+(11/4)*ln(x-2)
+(1/32)*ln(x-2)*2^(1/2)