Please help!

\[\int_{0}^{1}\dfrac{x^4+1}{x^6+1}dx=\dfrac{\pi}{3}\].Please prove the above statement.

Note by Adarsh Kumar
2 years, 10 months ago

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\[ \int_0^1 \dfrac{x^4+1}{x^6+1} = \int_0^1 \dfrac{x^4+1}{(x^2+1)(x^4 - x^2 + 1)} \]

\[= \int_0^1 \dfrac{x^4-x^2+1}{(x^2+1)(x^4 - x^2 + 1)} + \int_0^1 \dfrac{x^2}{(x^2+1)(x^4 - x^2 + 1)}\]

\[ = \int_0^1 \dfrac{1}{x^2+1} + \int_0^1 \dfrac{x^2}{x^6+1} \]


With \(\displaystyle \int_0^1 \dfrac{x^2}{x^6+1} dx \), substitute \(x^3 = u \Rightarrow \ 3x^2 \ dx = du \) to make it as \(\displaystyle \int_0^1 \dfrac{du}{3(u^2+1)} \).

The rest must be a baby's play for you I guess!

Satyajit Mohanty - 2 years, 10 months ago

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Nice solution !

Rajdeep Dhingra - 2 years, 10 months ago

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Many thanks,bhaiya.I am grateful.

Adarsh Kumar - 2 years, 10 months ago

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Well, while integrating Rational Functions of the form \(\dfrac{P(x)}{Q(x)}\), my first strategy is always to compare the degrees of \(P(x)\) and \(Q(x)\). If \(deg(P(x)) < deg(Q(x))\), I always invoke partial fractions. If \(deg(P(x)) > deg(Q(x) \), turn it into the form \(\dfrac{P(x)}{Q(x)} = R(x) + \dfrac{S(x)}{Q(x)} \) where \(deg(S(x)) < deg(Q(x))\), and then I invoke partial fractions!

Satyajit Mohanty - 2 years, 10 months ago

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@Satyajit Mohanty A lot of problems can be cracked by this statergy :)

Satyajit Mohanty - 2 years, 10 months ago

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