Question

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?

Answer 1

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.!