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valute ∫(1/((x−1)^2 ))(/((x^2 +4)))dx

$${valute}\:\int\frac{\mathrm{1}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }\frac{}{\left({x}^{\mathrm{2}} +\mathrm{4}\right)}{dx} \\ $$

Question Number 25479 Answers: 1 Comments: 0

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((1+x)/2)=x−3+(5/x) LCM=( 2)(x) multiply each step by 2x (2)(x)(((1+x)/2))=(2)(x)(x−3)+(2)(x)((5/x)) (x)(1+x)=2x(x−3)+(2)(5) x+x^2 =2x^2 −6x+10 move all equations to LHS 2x^2 −x^2 −6x−x+10=0 x^2 −5x+10=0 x=((5±(√(25−40)))/2)=((5±(√(−15)))/2)=(1/2)(5±i(√(15)))

$$\frac{\mathrm{1}+{x}}{\mathrm{2}}={x}−\mathrm{3}+\frac{\mathrm{5}}{{x}} \\ $$$${LCM}=\left(\:\mathrm{2}\right)\left({x}\right) \\ $$$${multiply}\:{each}\:{step}\:{by}\:\mathrm{2}{x} \\ $$$$ \\ $$$$\left(\mathrm{2}\right)\left({x}\right)\left(\frac{\mathrm{1}+{x}}{\mathrm{2}}\right)=\left(\mathrm{2}\right)\left({x}\right)\left({x}−\mathrm{3}\right)+\left(\mathrm{2}\right)\left({x}\right)\left(\frac{\mathrm{5}}{{x}}\right) \\ $$$$ \\ $$$$\left({x}\right)\left(\mathrm{1}+{x}\right)=\mathrm{2}{x}\left({x}−\mathrm{3}\right)+\left(\mathrm{2}\right)\left(\mathrm{5}\right)\:\:\:\:\:\:\:\:\:\:\: \\ $$$${x}+{x}^{\mathrm{2}} =\mathrm{2}{x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{10} \\ $$$${move}\:{all}\:{equations}\:{to}\:{LHS} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} −{x}^{\mathrm{2}} −\mathrm{6}{x}−{x}+\mathrm{10}=\mathrm{0} \\ $$$${x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{10}=\mathrm{0} \\ $$$${x}=\frac{\mathrm{5}\pm\sqrt{\mathrm{25}−\mathrm{40}}}{\mathrm{2}}=\frac{\mathrm{5}\pm\sqrt{−\mathrm{15}}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{5}\pm{i}\sqrt{\mathrm{15}}\right) \\ $$

Question Number 25476 Answers: 1 Comments: 0

convert 1234.24_8 to base 10

$${convert}\:\mathrm{1234}.\mathrm{24}_{\mathrm{8}} \:{to}\:{base}\:\mathrm{10} \\ $$

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slolve the definite integral∫_1^ ^2 (1/x)dxusingg> ng trapezoidal rule with 4 sub intervals hencefind an approximate value of ln 2.

$${slolve}\:{the}\:{definite}\:{integral}\int_{\mathrm{1}^{} } ^{\mathrm{2}} \frac{\mathrm{1}}{{x}}{dxusingg}>\:\:\:{ng}\: \\ $$$${trapezoidal}\:{rule}\:{with}\:\mathrm{4}\:{sub}\:{intervals}\: \\ $$$${hencefind}\:{an}\:{approximate}\:{value}\:{of}\: \\ $$$${ln}\:\mathrm{2}. \\ $$

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Question Number 26949 Answers: 0 Comments: 2

∫_0 ^1 ∫_0 ^1 (1/(1 + xy)) dx dy

$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\frac{\mathrm{1}}{\mathrm{1}\:+\:{xy}}\:{dx}\:{dy} \\ $$

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in what ratio in which y−x+2=0 divides the line joining (3,−1) and (8,9).

$${in}\:{what}\:{ratio}\:{in}\:{which}\:{y}−{x}+\mathrm{2}=\mathrm{0}\:{divides}\:{the}\:{line}\:{joining}\:\left(\mathrm{3},−\mathrm{1}\right)\:{and}\:\left(\mathrm{8},\mathrm{9}\right). \\ $$$$ \\ $$

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