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

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?

1 Answer
Jun 5, 2018

Kindly refer to the Discussion in Explanation.

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.!