Please help!

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

#Integration #DefiniteIntegral

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
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`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
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Comments

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

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

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

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 - 5 years, 10 months ago

Nice solution !

Rajdeep Dhingra - 5 years, 10 months ago

Many thanks,bhaiya.I am grateful.

Adarsh Kumar - 5 years, 10 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!

Satyajit Mohanty - 5 years, 10 months ago

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

Satyajit Mohanty - 5 years, 10 months ago

@Rajdeep Dhingra @Satyajit Mohanty @Aditya Kumar

Adarsh Kumar - 5 years, 10 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$