find the value of ∫_0 ^1 (√(x^4 −5x^2 +4)) dx? solution please.


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Question Number 142255 by gsk2684 last updated on 28/May/21

$find the value of \int_{0}^{1} \sqrt{x^{4} - {5 x}^{2} + 4} dx ?$ $solution please .$
	Answered by mindispower last updated on 28/May/21

	$b y p a r t = {[x \sqrt{x^{4} - 5 x^{2} + 4}]}_{0}^{1} - \int_{0}^{1} \frac{2 x^{4} - 5 x^{2}}{\sqrt{x^{4} - 5 x^{2} + 4}} d x$ $= - 2 \int_{0}^{1} \sqrt{x^{4} - 5 x^{2} + 4} + \int_{0}^{1} \frac{- 5 x^{2} + 8}{\sqrt{x^{4} - 5 x^{2} + 4}} d x$ $\int_{0}^{1} \sqrt{x^{4} + 5 x^{2} + 4} d x = \frac{1}{3} \int_{0}^{1} \frac{- 5 x^{2} + 8}{\sqrt{x^{4} - 5 x^{2} + 4}} d x$ $x^{4} - 5 x^{2} + 4 = (1 - x^{2}) (4 - x^{2})$ $\frac{1}{3} \int_{0}^{1} \frac{5 (4 - x^{2}) - 12}{\sqrt{x^{2} - 5 x^{2} + 4}} d x$ $= \frac{5}{3} \int_{0}^{1} \frac{\sqrt{4 - x^{2}}}{\sqrt{1 - x^{2}}} d x - 4 \int_{0}^{1} \frac{d x}{\sqrt{(1 - x^{2}) (4 - x^{2})}}$ $= \frac{10}{3} \int_{0}^{1} \frac{\sqrt{1 - \frac{1}{4} x^{2}}}{\sqrt{1 - x^{2}}} - 2 \int_{0}^{1} \frac{d x}{\sqrt{(1 - x^{2}) (1 - \frac{x^{2}}{4})}}$ $E (k^{2}) = \int_{0}^{1} \frac{\sqrt{1 - k^{2} x^{2}}}{\sqrt{1 - x^{2}}} d x 2 n d e l e p t i c i n t e g r a l$ $K (k^{2}) = \int_{0}^{1} \frac{d x}{\sqrt{1 - x^{2} . \sqrt{1 - k^{2} x^{2}}}} 1 s t k i n d e l e p t i c i n t e g r a l$ $w e G e t \int_{0}^{1} \frac{d x}{\sqrt{x^{4} - 5 x^{2} + 4}} = \frac{10}{3} E (\frac{1}{4}) - 2 K (\frac{1}{4})$
	Commented by MJS_new last updated on 29/May/21

	$great!$
	Commented by mindispower last updated on 29/May/21

	$t h a n x s i r$
	Commented by gsk2684 last updated on 31/May/21

	$thank you$
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