₁² \;x^4-1/x^6-1 dx= [AP EAPCET 18 May 2024 Shift 1]

Correct Answer is : 13Tan−1(32)\;{1}{{3}} {Tan}^{-1} ≤ft( {{3}}{2} )31Tan−1(23)

\;{1}{{3}} {Tan}^{-1} ≤ft( {{3}}{2} )

\;\;{1}{{2}} {Tan}^{-1} ≤ft( {2}{{3}} )

Correct Answer is : 13Tan−1(32)\;{1}{{3}} {Tan}^{-1} ≤ft( {{3}}{2} )31Tan−1(23)

Test Index

Definite Integration

Section: Mathematics

Question : 53 of 54

Marks: +1, -0

\int\limits_{1}^{2} \;\frac{x^4-1}{x^6-1} dx=

Solution:

\;\;\text{Let } I=\int\limits_{1}^{2} \;\frac{x^4-1}{x^6-1} dx

\;=\int\limits_{1}^{2} \;\frac{(x^2-1)(x^2+1)}{(x^2-1)(x^4+x^2+1)} dx

\;=\int\limits_{1}^{2} \;\frac{x^2+1}{x^4+x^2+1} dx=\int\limits_{1}^{2} \left( \frac{1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}+1} \right) dx

\;=\int\limits_{1}^{2} \left( \frac{1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}-2+3} \right) dx

\;=\int\limits_{1}^{2} \;\frac{1+\frac{1}{x^2}}{\left(x-\frac{1}{x}\right)^2+3} dx

Let

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

When

x \rightarrow 1

, then

t \rightarrow 0

When

x \rightarrow 2

, then

t \rightarrow \frac{3}{2}

I\;=\int\limits_{0}^{3/2} \;\frac{1}{t^2+3} dt

\;=\;\frac{1}{\sqrt{3}} \left[ \tan^{-1} \frac{1}{\sqrt{3}} \right]_{0}^{3/2}

\;=\;\frac{1}{\sqrt{3}} \tan^{-1} \frac{3}{2\sqrt{3}}

\;=\;\frac{1}{\sqrt{3}} \tan^{-1} \frac{\sqrt{3}}{2}

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