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Question Number 103511 by bemath last updated on 15/Jul/20

∫ (dx/((√x) ((x)^(1/4) +1))) =__ (a) −((9 (x)^(1/4) +1)/(18((x)^(1/4) +1)^9 )) + c (b) ((9 (x)^(1/4) +1)/(18((x)^(1/4) +1)^9 )) +c (c) −((9 (x)^(1/4) −1)/(18((x)^(1/4) +1)^9 )) +c (d) ((9 (x)^(1/4) +1)/(8((x)^(1/4) +1)^9 )) + c

$$\int\:\frac{{dx}}{\sqrt{{x}}\:\left(\sqrt[{\mathrm{4}}]{{x}}\:+\mathrm{1}\right)}\:=\_\_ \\ $$$$\left({a}\right)\:−\frac{\mathrm{9}\:\sqrt[{\mathrm{4}}]{{x}}\:+\mathrm{1}}{\mathrm{18}\left(\sqrt[{\mathrm{4}}]{{x}}\:+\mathrm{1}\right)^{\mathrm{9}} }\:+\:{c}\: \\ $$$$\left({b}\right)\:\frac{\mathrm{9}\:\sqrt[{\mathrm{4}}]{{x}}\:+\mathrm{1}}{\mathrm{18}\left(\sqrt[{\mathrm{4}}]{{x}}+\mathrm{1}\right)^{\mathrm{9}} }\:+{c} \\ $$$$\left({c}\right)\:−\frac{\mathrm{9}\:\sqrt[{\mathrm{4}}]{{x}}\:−\mathrm{1}}{\mathrm{18}\left(\sqrt[{\mathrm{4}}]{{x}}\:+\mathrm{1}\right)^{\mathrm{9}} }\:+{c} \\ $$$$\left({d}\right)\:\frac{\mathrm{9}\:\sqrt[{\mathrm{4}}]{{x}}+\mathrm{1}}{\mathrm{8}\left(\sqrt[{\mathrm{4}}]{{x}}\:+\mathrm{1}\right)^{\mathrm{9}} }\:+\:{c} \\ $$

Commented by bobhans last updated on 15/Jul/20

nothing answer

$${nothing}\:{answer} \\ $$

Answered by bramlex last updated on 15/Jul/20

let t = 1+(x)^(1/4) ⇒(x)^(1/4) = t−1 x= (t−1)^4 ∧ (√x) = (t−1)^2 dx = 4(t−1)^3 dt I= ∫ ((4(t−1)^3 dt)/((t−1)^2 .t ))= 4∫ ((t−1)/t) dt = 4∫(1−(1/t)) dt = 4t − 4ln∣t∣ +c = 4(1+(x)^(1/4) )−4ln∣1+(x)^(1/4) ∣ + c

$$\mathrm{let}\:\mathrm{t}\:=\:\mathrm{1}+\sqrt[{\mathrm{4}}]{\mathrm{x}}\:\Rightarrow\sqrt[{\mathrm{4}}]{\mathrm{x}}\:=\:\mathrm{t}−\mathrm{1} \\ $$$$\mathrm{x}=\:\left(\mathrm{t}−\mathrm{1}\right)^{\mathrm{4}} \:\wedge\:\sqrt{\mathrm{x}}\:=\:\left(\mathrm{t}−\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\mathrm{dx}\:=\:\mathrm{4}\left(\mathrm{t}−\mathrm{1}\right)^{\mathrm{3}} \:\mathrm{dt}\: \\ $$$$\mathrm{I}=\:\int\:\frac{\mathrm{4}\left(\mathrm{t}−\mathrm{1}\right)^{\mathrm{3}} \:\mathrm{dt}}{\left(\mathrm{t}−\mathrm{1}\right)^{\mathrm{2}} .\mathrm{t}\:}=\:\mathrm{4}\int\:\frac{\mathrm{t}−\mathrm{1}}{\mathrm{t}}\:\mathrm{dt} \\ $$$$=\:\mathrm{4}\int\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{t}}\right)\:\mathrm{dt}\: \\ $$$$=\:\mathrm{4t}\:−\:\mathrm{4ln}\mid\mathrm{t}\mid\:+\mathrm{c}\: \\ $$$$=\:\mathrm{4}\left(\mathrm{1}+\sqrt[{\mathrm{4}}]{\mathrm{x}}\right)−\mathrm{4ln}\mid\mathrm{1}+\sqrt[{\mathrm{4}}]{\mathrm{x}}\:\mid\:+\:\mathrm{c}\: \\ $$

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