{x²-1}{x³{2x^4-2x²+1}} dx is equal to :

Correct Answer is : 2x4−2x2+12x2{{2x^4-2x²+1}}{2x²2x22x4−2x2+1 + C

Correct Answer is : 2x4−2x2+12x2{{2x^4-2x²+1}}{2x²2x22x4−2x2+1 + C

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NIMCET 2016 Question Paper with solutions

Question : 63 of 120

Marks: +1, -0

\int\frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}}

dx is equal to :

Solution:

Here

l=\int\frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}}dx

[On dividing numerators and denominators by

x^5

]

\Rightarrow l=\int\frac{\frac{x^2-1}{x^5}}{\frac{x^3\sqrt{2x^4-2x^2+1}}{x^5}}dx

=\int\frac{\frac{1}{x^3}-\frac{1}{x^5}}{\frac{x^2\sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}}{x^2}}dx

=\int\frac{\frac{1}{x^3}-\frac{1}{x^5}}{\sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}}dx

Let

\left(2-\frac{2}{x^2}+\frac{1}{x^4}\right)=t

\Rightarrow\left(\frac{4}{x^3}-\frac{4}{x^5}\right)dx=dt

\therefore l=\frac{1}{4}\int\frac{dt}{\sqrt{C}}

=\frac{1}{4}\int t^{-1/2}dt

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

[Here C is integration constant]

=\frac{1}{4}\times2t^{1/2}+C

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

On substituting the value oft, then

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

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

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

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

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