\int {x^2-1}/{x^3\sqrt{2x^4-2x^2+1} dx is equal to :

Correct Answer is : 2x4−2x2+12x2 + C

\sqrt\frac{2x^4-2x^2+1}{2x^2} + C

NIMCET 2016 Question Paper with solutions

Question : 63 of 120

Marks: +1, -0

∫ {x^2-1}/{x^3√{2x^4-2x^2+1}}

dx is equal to :

Solution:

Here

l=∫{x^2-1}/{x^3√{2x^4-2x^2+1}}dx

[On dividing numerators and denominators by

x^5

]

⇒l=∫{{x^2-1}/x^5}/{x^3√{2x^4-2x^2+1}}/x^5dx

=∫{1/x^3-1/x^5}/{x^2√{2-2/x^2+1/x^4}}/x^2dx

=∫{1/x^3-1/x^5}/{√{2-2/x^2+1/x^4}}dx

Let

(2-2/x^2+1/x^4)=t

⇒(4/x^3-4/x^5)dx=dt

∴l=1/4∫{dt}/√C

=1/4∫t^{-1/2}dt

=1/4t^{1/2}/1/2+C

[Here C is integration constant]

=1/4×2t^{1/2}+C

=1/2t^{1/2}+C

On substituting the value oft, then

l=\;\;{1}/{2}√{2-\;\;{2}/{x^{2}}+\;\;{1}/{x^{4}}}+C

=\;\;{1}/{2}√{\;\;{2x^{4}-2x^{2}+1}/{x^{4}}}+C

=\;\;{1}/{2}\;\;{√{2x^{4}-2x^{2}+1}}/{x^{2}}+C

=\;\;{√{2x^{4}-2x^{2}+1}}/{2x^{2}}+C

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