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Question Number 44712 by peter frank last updated on 03/Oct/18

Answered by tanmay.chaudhury50@gmail.com last updated on 03/Oct/18

1)∫(x^(n−1) /(x^n (1+x^n )))dx t=x^n dt=nx^(n−1) dx (1/n)∫(dt/(t(1+t))) (1/n)[∫(((1+t)−t)/(t(1+t)))dt (1/n)[∫(dt/t)−∫(dt/(1+t))] (1/n)[ln((t/(1+t)))]+c (1/n)[ln((x^n /(1+x^n )))]+c

$$\left.\mathrm{1}\right)\int\frac{{x}^{{n}−\mathrm{1}} }{{x}^{{n}} \left(\mathrm{1}+{x}^{{n}} \right)}{dx} \\ $$$${t}={x}^{{n}} \:\:\:{dt}={nx}^{{n}−\mathrm{1}} {dx} \\ $$$$\frac{\mathrm{1}}{{n}}\int\frac{{dt}}{{t}\left(\mathrm{1}+{t}\right)} \\ $$$$\frac{\mathrm{1}}{{n}}\left[\int\frac{\left(\mathrm{1}+{t}\right)−{t}}{{t}\left(\mathrm{1}+{t}\right)}{dt}\right. \\ $$$$\frac{\mathrm{1}}{{n}}\left[\int\frac{{dt}}{{t}}−\int\frac{{dt}}{\mathrm{1}+{t}}\right] \\ $$$$\frac{\mathrm{1}}{{n}}\left[{ln}\left(\frac{{t}}{\mathrm{1}+{t}}\right)\right]+{c} \\ $$$$\frac{\mathrm{1}}{{n}}\left[{ln}\left(\frac{{x}^{{n}} }{\mathrm{1}+{x}^{{n}} }\right)\right]+{c} \\ $$

Commented by peter frank last updated on 03/Oct/18

thank you

$$\mathrm{thank}\:\mathrm{you} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 03/Oct/18

3)x+α=t ∫((sin(t−2α))/(sint))dt ∫((sintcos2α−costsin2α)/(sint))dt cos2α∫dt−sin2α∫((cost)/(sint))dt cos2α(t)−sin2α lnsint+c (x+α)cos2α−sin2α lnsin(x+α)+c

$$\left.\mathrm{3}\right){x}+\alpha={t} \\ $$$$\int\frac{{sin}\left({t}−\mathrm{2}\alpha\right)}{{sint}}{dt} \\ $$$$\int\frac{{sintcos}\mathrm{2}\alpha−{costsin}\mathrm{2}\alpha}{{sint}}{dt} \\ $$$${cos}\mathrm{2}\alpha\int{dt}−{sin}\mathrm{2}\alpha\int\frac{{cost}}{{sint}}{dt} \\ $$$${cos}\mathrm{2}\alpha\left({t}\right)−{sin}\mathrm{2}\alpha\:{lnsint}+{c} \\ $$$$\left({x}+\alpha\right){cos}\mathrm{2}\alpha−{sin}\mathrm{2}\alpha\:{lnsin}\left({x}+\alpha\right)+{c} \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 03/Oct/18

pls see it and chek the answer you provided along with question...

$${pls}\:{see}\:{it}\:{and}\:{chek}\:{the}\:{answer}\:{you}\:{provided} \\ $$$${along}\:{with}\:{question}... \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 03/Oct/18

i think attached answer not correvct...

$${i}\:{think}\:{attached}\:{answer}\:{not}\:{correvct}... \\ $$

Commented by peter frank last updated on 03/Oct/18

true sir very sorry

$$\mathrm{true}\:\mathrm{sir}\:\mathrm{very}\:\mathrm{sorry} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 03/Oct/18

2)∫(((x^4 −x)^(1/4) )/x^5 )dx ∫(({x^4 (1−(1/x^3 ))}^(1/4) )/x^5 ) ∫(((1−(1/x^3 ))^(1/4) )/x^4 )dx t=1−(1/x^3 ) dt=0−(−3)x^(−4) dx dt=((3dx)/x^4 ) ∫t^(1/4) ×(dt/3) (1/3)×(t^((1/4)+1) /((1/4)+1))+c (4/(15))t^(5/4) +c (4/(15))×(1−(1/x^3 ))^(5/4) +c

$$\left.\mathrm{2}\right)\int\frac{\left({x}^{\mathrm{4}} −{x}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} }{{x}^{\mathrm{5}} }{dx} \\ $$$$\int\frac{\left\{{x}^{\mathrm{4}} \left(\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\right)\right\}^{\frac{\mathrm{1}}{\mathrm{4}}} }{{x}^{\mathrm{5}} } \\ $$$$\int\frac{\left(\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\right)^{\frac{\mathrm{1}}{\mathrm{4}}} }{{x}^{\mathrm{4}} }{dx} \\ $$$${t}=\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\:\:\:\:{dt}=\mathrm{0}−\left(−\mathrm{3}\right){x}^{−\mathrm{4}} {dx} \\ $$$${dt}=\frac{\mathrm{3}{dx}}{{x}^{\mathrm{4}} } \\ $$$$\int{t}^{\frac{\mathrm{1}}{\mathrm{4}}} ×\frac{{dt}}{\mathrm{3}} \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}×\frac{{t}^{\frac{\mathrm{1}}{\mathrm{4}}+\mathrm{1}} }{\frac{\mathrm{1}}{\mathrm{4}}+\mathrm{1}}+{c} \\ $$$$\frac{\mathrm{4}}{\mathrm{15}}{t}^{\frac{\mathrm{5}}{\mathrm{4}}} +{c}\:\:\: \\ $$$$\frac{\mathrm{4}}{\mathrm{15}}×\left(\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\right)^{\frac{\mathrm{5}}{\mathrm{4}}} \:+{c} \\ $$$$ \\ $$$$ \\ $$

Commented by peter frank last updated on 03/Oct/18

thank you

$$\mathrm{thank}\:\mathrm{you} \\ $$

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