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Question Number 53383 Answers: 1 Comments: 1

If ∫(1/((sin x+4)(sin x−1)))dx = A(1/(tan (x/2)−1))+B tan^(−1) (f(x))+C, then

$$\mathrm{If}\:\int\frac{\mathrm{1}}{\left(\mathrm{sin}\:{x}+\mathrm{4}\right)\left(\mathrm{sin}\:{x}−\mathrm{1}\right)}{dx} \\ $$$$\:\:\:\:\:\:\:\:\:=\:{A}\frac{\mathrm{1}}{\mathrm{tan}\:\frac{{x}}{\mathrm{2}}−\mathrm{1}}+{B}\:\mathrm{tan}^{−\mathrm{1}} \left({f}\left({x}\right)\right)+{C},\:\mathrm{then} \\ $$

Question Number 53382 Answers: 1 Comments: 0

∫ cos^3 x e^(log (sin x)) dx =

$$\int\:\mathrm{cos}^{\mathrm{3}} {x}\:{e}^{\mathrm{log}\:\left(\mathrm{sin}\:{x}\right)} {dx}\:= \\ $$

Question Number 53381 Answers: 1 Comments: 0

∫ (1/([(x−1)^3 (x+2)^5 ]^(1/4) )) dx =

$$\int\:\frac{\mathrm{1}}{\left[\left({x}−\mathrm{1}\right)^{\mathrm{3}} \left({x}+\mathrm{2}\right)^{\mathrm{5}} \right]^{\mathrm{1}/\mathrm{4}} }\:{dx}\:= \\ $$

Question Number 53378 Answers: 1 Comments: 0

if u=e^(xyz) then u_(xyx) =? a)u((xyz)^2 +3xyz+1) b)u(3(xyz)^2 +1) c)u((xyz)^2 +2yz+1) please help

$${if}\:{u}={e}^{{xyz}} \:{then}\:{u}_{{xyx}} =? \\ $$$$\left.{a}\left.\right){u}\left(\left({xyz}\right)^{\mathrm{2}} +\mathrm{3}{xyz}+\mathrm{1}\right)\:{b}\right){u}\left(\mathrm{3}\left({xyz}\right)^{\mathrm{2}} +\mathrm{1}\right) \\ $$$$\left.{c}\right){u}\left(\left({xyz}\right)^{\mathrm{2}} +\mathrm{2}{yz}+\mathrm{1}\right) \\ $$$$ \\ $$$${please}\:{help} \\ $$

Question Number 53376 Answers: 2 Comments: 1

Question Number 53359 Answers: 2 Comments: 0

∫ ((x^3 −1)/(x^3 +x)) dx =

$$\int\:\:\frac{{x}^{\mathrm{3}} −\mathrm{1}}{{x}^{\mathrm{3}} +{x}}\:{dx}\:= \\ $$

Question Number 53358 Answers: 1 Comments: 0

∫ (1/(√(sin^3 x cos x))) dx =

$$\int\:\:\frac{\mathrm{1}}{\sqrt{\mathrm{sin}^{\mathrm{3}} {x}\:\mathrm{cos}\:{x}}}\:{dx}\:= \\ $$

Question Number 53357 Answers: 1 Comments: 0

∫ (((x^4 −x)^(1/4) )/x^5 ) dx =

$$\int\:\:\frac{\left({x}^{\mathrm{4}} −{x}\right)^{\mathrm{1}/\mathrm{4}} }{{x}^{\mathrm{5}} }\:{dx}\:= \\ $$

Question Number 53353 Answers: 1 Comments: 0

Question Number 53351 Answers: 0 Comments: 1

Question Number 53349 Answers: 0 Comments: 2

Question Number 53342 Answers: 0 Comments: 4

Question Number 53340 Answers: 1 Comments: 0

Question Number 53330 Answers: 0 Comments: 12

Question Number 53328 Answers: 1 Comments: 2

Question Number 53325 Answers: 0 Comments: 10

Question Number 53324 Answers: 1 Comments: 0

If 4a + 5b + 9c=36 and 7a + 9b + 17c=66, then a+b+c=_____.

$$\mathrm{If}\:\mathrm{4}{a}\:+\:\mathrm{5}{b}\:+\:\mathrm{9}{c}=\mathrm{36}\:\mathrm{and}\:\mathrm{7}{a}\:+\:\mathrm{9}{b}\:+\:\mathrm{17}{c}=\mathrm{66}, \\ $$$$\mathrm{then}\:{a}+{b}+{c}=\_\_\_\_\_. \\ $$

Question Number 53323 Answers: 1 Comments: 0

Question Number 53318 Answers: 1 Comments: 1

Question Number 53311 Answers: 1 Comments: 1

If ∫ ((4e^x +6e^(−x) )/(9e^x −4e^(−x) )) dx=Ax+B log(9e^(2x) −4)+C then A=... B=... C=...

$$\mathrm{If}\:\int\:\frac{\mathrm{4}{e}^{{x}} +\mathrm{6}{e}^{−{x}} }{\mathrm{9}{e}^{{x}} −\mathrm{4}{e}^{−{x}} }\:{dx}={Ax}+{B}\:\mathrm{log}\left(\mathrm{9}{e}^{\mathrm{2}{x}} −\mathrm{4}\right)+{C} \\ $$$$\mathrm{then} \\ $$$${A}=... \\ $$$${B}=... \\ $$$${C}=... \\ $$

Question Number 53295 Answers: 1 Comments: 1

∫_0 ^(π/2) (1/(2+cos x)) dx=...

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}}{\mathrm{2}+\mathrm{cos}\:{x}}\:{dx}=... \\ $$

Question Number 53294 Answers: 1 Comments: 0

∫_(−1/2) ^(1/2) [(((x+1)/(x−1)))^2 +(((x−1)/(x+1)))^2 −2]^(1/2) dx=...

$$\int_{−\mathrm{1}/\mathrm{2}} ^{\mathrm{1}/\mathrm{2}} \left[\left(\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}\right)^{\mathrm{2}} +\left(\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}\right)^{\mathrm{2}} −\mathrm{2}\right]^{\mathrm{1}/\mathrm{2}} {dx}=... \\ $$

Question Number 53293 Answers: 1 Comments: 1

∫_(−1/2) ^(1/2) ∣xcos ((πx)/2)∣ dx=...

$$\int_{−\mathrm{1}/\mathrm{2}} ^{\mathrm{1}/\mathrm{2}} \mid{x}\mathrm{cos}\:\frac{\pi{x}}{\mathrm{2}}\mid\:{dx}=... \\ $$

Question Number 53292 Answers: 1 Comments: 1

∫_0 ^1 e^x^2 dx=..

$$\int_{\mathrm{0}} ^{\mathrm{1}} {e}^{{x}^{\mathrm{2}} } {dx}=.. \\ $$

Question Number 53285 Answers: 0 Comments: 0

let I_λ =∫_0 ^π ((xdx)/(cos^2 x +λ^2 sin^2 x)) with λ real 1) find the value of I_λ 2) calculate ∫_0 ^π ((xdx)/(a^2 cos^2 x +b^2 sin^2 x)) with a and b reals.

$${let}\:{I}_{\lambda} \:=\int_{\mathrm{0}} ^{\pi} \:\:\:\frac{{xdx}}{{cos}^{\mathrm{2}} {x}\:+\lambda^{\mathrm{2}} {sin}^{\mathrm{2}} {x}}\:\:{with}\:\lambda\:{real} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{the}\:{value}\:{of}\:{I}_{\lambda} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\pi} \:\:\frac{{xdx}}{{a}^{\mathrm{2}} {cos}^{\mathrm{2}} {x}\:+{b}^{\mathrm{2}} {sin}^{\mathrm{2}} {x}}\:{with}\:{a}\:{and}\:{b}\:{reals}. \\ $$

Question Number 53284 Answers: 0 Comments: 2

find f(x)=∫_0 ^∞ ((arctan(xt))/(1+t^2 ))dt with x real .

$${find}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{{arctan}\left({xt}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\:\:\:\:{with}\:{x}\:{real}\:. \\ $$

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