xdx-x-4-please-

Question Number 206721 by ajfour last updated on 23/Apr/24

$$\int\frac{{xdx}}{{x}+\mathrm{4}}=?\:\:\:\:\:\:\:{please} \\ $$

Answered by A5T last updated on 23/Apr/24

(x/(x+4))=((x+4−4)/(x+4))=1−(4/(x+4)) ⇒∫(x/(x+4))dx=∫1dx−4∫(1/(x+4))dx =x−4ln∣x+4∣+c

$$\frac{{x}}{{x}+\mathrm{4}}=\frac{{x}+\mathrm{4}−\mathrm{4}}{{x}+\mathrm{4}}=\mathrm{1}−\frac{\mathrm{4}}{{x}+\mathrm{4}} \\ $$$$\Rightarrow\int\frac{{x}}{{x}+\mathrm{4}}{dx}=\int\mathrm{1}{dx}−\mathrm{4}\int\frac{\mathrm{1}}{{x}+\mathrm{4}}{dx} \\ $$$$={x}−\mathrm{4}{ln}\mid{x}+\mathrm{4}\mid+{c} \\ $$

Commented by ajfour last updated on 23/Apr/24

$${Oh}\:{yes}!\:{thank}\:{you}.\:{this}\:{is}\:{the}\:{answer}. \\ $$

Commented by necx122 last updated on 23/Apr/24

It's funny how after after a long while we see Ajfour's post and it's a quite simple integral relative to what we know him for. Welcome back our prof.��

Commented by mr W last updated on 23/Apr/24

$${i}\:{guess}\:{it}'{s}\:{not}\:{him}\:{personally}. \\ $$

Commented by ajfour last updated on 23/Apr/24

yeah me, bit irritated yet amused by life. It will get okay, assure you.

Commented by mr W last updated on 23/Apr/24

$${then}\:{welcome}\:{back}\:{sir}! \\ $$

Answered by Ghisom last updated on 23/Apr/24

∫(x/(x+4))dx= [t=((√(x+4))/2) → dx=4(√(x+4))dt] =8∫(t−(1/t))dt=4t^2 −8ln t = =x−4ln ∣x+4∣ +C

$$\int\frac{{x}}{{x}+\mathrm{4}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\frac{\sqrt{{x}+\mathrm{4}}}{\mathrm{2}}\:\rightarrow\:{dx}=\mathrm{4}\sqrt{{x}+\mathrm{4}}{dt}\right] \\ $$$$=\mathrm{8}\int\left({t}−\frac{\mathrm{1}}{{t}}\right){dt}=\mathrm{4}{t}^{\mathrm{2}} −\mathrm{8ln}\:{t}\:= \\ $$$$={x}−\mathrm{4ln}\:\mid{x}+\mathrm{4}\mid\:+{C} \\ $$

Commented by ajfour last updated on 23/Apr/24

$${wow}! \\ $$

Answered by BaliramKumar last updated on 24/Apr/24

put x = 4tan^2 y dx = 8tanysec^2 ydy ∫((4tan^2 y)/(4tan^2 y+4))∙8tanysec^2 ydy ∫((4tan^2 y)/(4sec^2 y))∙8tanysec^2 ydy 8∫tany(tan^2 y)dy = 8∫tany(sec^2 y−1)dy 8∫tanysec^2 ydy−8∫tanydy 8[((tan^2 y)/2)] − 8ln(secy) 4tan^2 y − 4ln(sec^2 y) = x − 4ln(((4sec^2 y)/4)) x − 4ln(((4tan^2 y + 4)/4)) = x − 4ln(((x + 4)/4)) x − 4ln∣x+4∣ + 4ln∣4∣ x − 4ln∣x+4∣ + C

$$\mathrm{put}\:\:\:\:\:{x}\:=\:\mathrm{4}{tan}^{\mathrm{2}} {y}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{dx}\:=\:\mathrm{8}{tanysec}^{\mathrm{2}} {ydy} \\ $$$$\int\frac{\mathrm{4tan}^{\mathrm{2}} \mathrm{y}}{\mathrm{4tan}^{\mathrm{2}} \mathrm{y}+\mathrm{4}}\centerdot\mathrm{8tanysec}^{\mathrm{2}} \mathrm{ydy} \\ $$$$\int\frac{\mathrm{4tan}^{\mathrm{2}} \mathrm{y}}{\mathrm{4sec}^{\mathrm{2}} \mathrm{y}}\centerdot\mathrm{8tanysec}^{\mathrm{2}} \mathrm{ydy} \\ $$$$\mathrm{8}\int\mathrm{tany}\left(\mathrm{tan}^{\mathrm{2}} \mathrm{y}\right)\mathrm{dy}\:=\:\mathrm{8}\int\mathrm{tany}\left(\mathrm{sec}^{\mathrm{2}} \mathrm{y}−\mathrm{1}\right)\mathrm{dy}\: \\ $$$$\mathrm{8}\int\mathrm{tanysec}^{\mathrm{2}} \mathrm{ydy}−\mathrm{8}\int\mathrm{tanydy}\: \\ $$$$\mathrm{8}\left[\frac{\mathrm{tan}^{\mathrm{2}} \mathrm{y}}{\mathrm{2}}\right]\:−\:\mathrm{8ln}\left(\mathrm{secy}\right) \\ $$$$\mathrm{4tan}^{\mathrm{2}} \mathrm{y}\:−\:\mathrm{4ln}\left(\mathrm{sec}^{\mathrm{2}} \mathrm{y}\right)\:=\:{x}\:−\:\mathrm{4ln}\left(\frac{\mathrm{4sec}^{\mathrm{2}} \mathrm{y}}{\mathrm{4}}\right) \\ $$$${x}\:−\:\mathrm{4ln}\left(\frac{\mathrm{4}{tan}^{\mathrm{2}} {y}\:+\:\mathrm{4}}{\mathrm{4}}\right)\:=\:{x}\:−\:\mathrm{4ln}\left(\frac{{x}\:+\:\mathrm{4}}{\mathrm{4}}\right) \\ $$$$\:{x}\:−\:\mathrm{4ln}\mid{x}+\mathrm{4}\mid\:+\:\mathrm{4}{ln}\mid\mathrm{4}\mid \\ $$$${x}\:−\:\mathrm{4}{ln}\mid{x}+\mathrm{4}\mid\:+\:\mathrm{C} \\ $$$$ \\ $$

Answered by peter frank last updated on 24/Apr/24

u=x+4 x=u−4 du=dx ∫((xdu)/u)=∫((u−4)/u)du=∫1du−∫(4/u)=u−4ln u ∫1du−∫(4/u)du=u−4ln u x+4−4ln (x+4)

$$\mathrm{u}=\mathrm{x}+\mathrm{4}\:\:\mathrm{x}=\mathrm{u}−\mathrm{4} \\ $$$$\mathrm{du}=\mathrm{dx} \\ $$$$\int\frac{\mathrm{xdu}}{\mathrm{u}}=\int\frac{\mathrm{u}−\mathrm{4}}{\mathrm{u}}\mathrm{du}=\int\mathrm{1du}−\int\frac{\mathrm{4}}{\mathrm{u}}=\mathrm{u}−\mathrm{4ln}\:\mathrm{u} \\ $$$$\int\mathrm{1du}−\int\frac{\mathrm{4}}{\mathrm{u}}\mathrm{du}=\mathrm{u}−\mathrm{4ln}\:\mathrm{u} \\ $$$$\mathrm{x}+\mathrm{4}−\mathrm{4ln}\:\left(\mathrm{x}+\mathrm{4}\right) \\ $$

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