What does #int(xdx)/(2x^4 - 3x^2-2# equal to?

Question

I got my answer as #log_e{[(x^2-2)/(2x^2+1)]^(1/10)}# while in the answer key, the correct answer has been given as #log_e{[(2x^2-4)/(2x^2+1)]^(1/10)}# . I went over my calculations again and again but still got the answer I previously mentioned.
I used the partial fractions technique to break the given function down and then went ahead doing integration.

Can someone please post their solution for this question here and give a confirmation on which one out of these two is actually the correct answer?

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Answer 1 · 2018-06-05T17:56:01

Kindly refer to the Discussion in Explanation.

Let, #I=int(xdx)/(2x^4-3x^2-2)=int(2xdx)/{(2(x^2-2))(2x^2+1)}#.

#:. I=int(2xdx)/{(2x^2-4)(2x^2+1)}#.

Subst.ing #x^2=t," so that, "2xdx=dt#, we have,

#I=intdt/{(2t-4)(2t+1)}#,

#=1/5int(5dt)/((2t-4)(2t+1))#.

Here, note that, #(2t+1)-(2t-4)=5#.

#:. I=1/5int(5dt)/((2t-4)(2t+1))#,

#=1/5int{(2t+1)-(2t-4)}/((2t-4)(2t+1))dt#,

#=1/5int{(2t+1)/((2t-4)(2t+1))-(2t-4)/((2t-4)(2t+1))}dt#,

#=1/5int{1/(2t-4)-1/(2t+1)}dt#,

#=1/5{(log_e|(2t-4)|)/2-(log_e|(2t+1)|)/2}#,

#=1/10log_e|((2t-4)/(2t+1))|#.

Since, #t=x^2, I=log_e|((2x^2-4)/(2x^2+1))|^(1/10)+C.#

By the way, your Answer is quite right, as can be seen from the

following :

As per the Answer Key, #I=log_e|((2x^2-4)/(2x^2+1))|^(1/10)+C#,

#=log_e|((2(x^2-2))/(2x^2+1))|^(1/10)+C#,

#=log_e|((2*(x^2-2))/(2x^2+1))|^(1/10)+C#,

#=log_e|(2^(1/10))*((x^2-2)/(2x^2+1))^(1/10)|+C#,

#=log_e 2^(1/10)+log_e|((x^2-2)/(2x^2+1))^(1/10)|+C#,

#=log_e|((x^2-2)/(2x^2+1))^(1/10)|+C_1," where, "C_1=C+log_e 2^(1/10)#,

as per your Answer!

Enjoy Maths.!