Integration on Partial Fraction

[Seimuna]

Question 1
(3x+7)/((x^4)-16)
the partial fraction i get is (13/31)(1/(x-2)) - (1/32)(1/(x+2)) - ((3x+7)/(8(x^2)+4)
then integration of the partial fraction i got 13/32 ln |x-2| - 1/32 ln|x+2| - 3/16 ln |(x^2)/4 + 1| + 7/32 tan^-1 (x/2)
but the answer given is 13/32 ln |x-2| - 1/32 ln|x+2| - 3/16 ln |(x^2)/4 + 1| + 7/16 tan^-1 (x/2)

Question 2
((x^3)-2x) / (2x^2-3x+2)
using long division i got x/2+3/4 +(3/4)(x+2)/(2(x^2)-3x+2)
the problem i facing now is integration of (3/4)(x+2)/(2(x^2)-3x+2)
the answer given is quite weird... (3/4) ln [(1/14){(4xSqrt[2]-3Sqrt[2])^2} + 1] - { [15 tan^-1 {(4xSqrt[2]-3Sqrt[2])/(Sqrt[14])} ] / 2Sqrt[7] }

Thanks in advance...

 

[Deleted member 4993]

Question 1
(3x+7)/((x^4)-16)
the partial fraction i get is (13/31)(1/(x-2)) - (1/32)(1/(x+2)) - ((3x+7)/(8(x^2)+4)
then integration of the partial fraction i got 13/32 ln |x-2| - 1/32 ln|x+2| - 3/16 ln |(x^2)/4 + 1| + 7/32 tan^-1 (x/2)
but the answer given is 13/32 ln |x-2| - 1/32 ln|x+2| - 3/16 ln |(x^2)/4 + 1| + 7/16 tan^-1 (x/2)

Question 2
((x^3)-2x) / (2x^2-3x+2)

using long division i got x/2+3/4 +(3/4)(x+2)/(2(x^2)-3x+2)
the problem i facing now is integration of (3/4)(x+2)/(2(x^2)-3x+2)

2x^2 - 3x -2 = 2[(x-3/2)^2 + {sqrt(23/32)}^2]

replace u = x-3/2 and a = sqrt(23/32)

2[(x-3/2)^2 + {sqrt(23/32)}^2] = 2[u^2 + a^2]

That should look familiar


the answer given is quite weird... (3/4) ln [(1/14){(4xSqrt[2]-3Sqrt[2])^2} + 1] - { [15 tan^-1 {(4xSqrt[2]-3Sqrt[2])/(Sqrt[14])} ] / 2Sqrt[7] }

Thanks in advance...