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laplace transform 2cos^3(x)

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`x`²	`x`^□	log_□	√☐	^□√☐	≤	≥	□□	·	÷	`x`^◦	π
(☐)^′	`ddx`	∂∂`x`	∫	∫_□^□	lim	∑	∞	`θ`	(`f` ◦ `g`)	`f`(`x`)

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∫ x²−25x²+4x−5 dx

Solution

x−4ln|x−1|+C

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∫ x²−25x²+4x−5 dx

以下の部分分数を得る: x²−25x²+4x−5 : 1−4x−1

=∫ 1−4x−1 dx

総和規則を適用する: ∫f(x)±g(x)dx=∫f(x)dx±∫g(x)dx

=∫ 1dx−∫ 4x−1 dx

∫ 1dx=x

∫ 4x−1 dx=4ln|x−1|

=x−4ln|x−1|

定数を解答に追加する

=x−4ln|x−1|+C

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こんにちは, について質問する ∫ x²−25x²+4x−5 dx または以下のオプションから選択してください

不定積分の解き方を説明する

−▭▭	<	7	8	9	÷	`AC`
+▭▭	>	4	5	6	×	☐☐☐
×▭▭	(	1	2	3	−	`x`
▭ \|▭	)	.	0	=	+	`y`

laplace transform 2cos^3(x)

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laplace transform 2cos^3(x)

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Solution steps

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