xdx-x-4-please-

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

\int \frac{x d x}{x + 4} = ? p l e a s e

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 + 4} = \frac{x + 4 - 4}{x + 4} = 1 - \frac{4}{x + 4}

\Rightarrow \int \frac{x}{x + 4} d x = \int 1 d x - 4 \int \frac{1}{x + 4} d x

= x - 4 l n ∣ x + 4 ∣ + c

Commented by ajfour last updated on 23/Apr/24

O h y e s! t h a n k y o u . t h i s i s t h e a n s w e r .

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 g u e s s {i t}^{'} s n o t h i m p e r s o n a l l y .

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

t h e n w e l c o m e b a c k s i r!

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 + 4} d x =

[t = \frac{\sqrt{x + 4}}{2} \to d x = 4 \sqrt{x + 4} d t]

= 8 \int (t - \frac{1}{t}) d t = 4 t^{2} - 8 \ln t =

= x - 4 \ln ∣ x + 4 ∣ + C

Commented by ajfour last updated on 23/Apr/24

w o w!

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

put x = 4 {t a n}^{2} y d x = 8 {t a n y s e c}^{2} y d y

\int \frac{{4 \tan}^{2} y}{{4 \tan}^{2} y + 4} \cdot {8 tanysec}^{2} ydy

\int \frac{{4 \tan}^{2} y}{{4 \sec}^{2} y} \cdot {8 tanysec}^{2} ydy

8 \int tany (\tan^{2} y) dy = 8 \int tany (\sec^{2} y - 1) dy

8 \int {tanysec}^{2} ydy - 8 \int tanydy

8 [\frac{\tan^{2} y}{2}] - 8 \ln (secy)

{4 \tan}^{2} y - 4 \ln (\sec^{2} y) = x - 4 \ln (\frac{{4 \sec}^{2} y}{4})

x - 4 \ln (\frac{4 {t a n}^{2} y + 4}{4}) = x - 4 \ln (\frac{x + 4}{4})

x - 4 \ln ∣ x + 4 ∣ + 4 l n ∣ 4 ∣

x - 4 l n ∣ x + 4 ∣ + 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)

u = x + 4 x = u - 4

du = dx

\int \frac{xdu}{u} = \int \frac{u - 4}{u} du = \int 1 du - \int \frac{4}{u} = u - 4 \ln u

\int 1 du - \int \frac{4}{u} du = u - 4 \ln u

x + 4 - 4 \ln (x + 4)

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