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

1 year, 11 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! · 1 year, 11 months ago

Nice solution ! · 1 year, 11 months ago

Many thanks,bhaiya.I am grateful. · 1 year, 11 months ago

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! · 1 year, 11 months ago

A lot of problems can be cracked by this statergy :) · 1 year, 11 months ago