∫((xdx)/(x+4))=? please


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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

	$\frac{x}{x + 4} = \frac{x + 4 - 4}{x + 4} = 1 - \frac{4}{x + 4}$ $\Rightarrow \int \frac{x}{x + 4} d x = \int 1 d x - 4 \int \frac{1}{x + 4} d x$ $= x - 4 l n ∣ x + 4 ∣ + c$
	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

	$\int \frac{x}{x + 4} d x =$ $[t = \frac{\sqrt{x + 4}}{2} \to d x = 4 \sqrt{x + 4} d t]$ $= 8 \int (t - \frac{1}{t}) d t = 4 t^{2} - 8 \ln t =$ $= x - 4 \ln ∣ x + 4 ∣ + C$
	Commented by ajfour last updated on 23/Apr/24

	$w o w!$
	Answered by BaliramKumar last updated on 24/Apr/24

	$put x = 4 {t a n}^{2} y d x = 8 {t a n y s e c}^{2} y d y$ $\int \frac{{4 \tan}^{2} y}{{4 \tan}^{2} y + 4} \cdot {8 tanysec}^{2} ydy$ $\int \frac{{4 \tan}^{2} y}{{4 \sec}^{2} y} \cdot {8 tanysec}^{2} ydy$ $8 \int tany (\tan^{2} y) dy = 8 \int tany (\sec^{2} y - 1) dy$ $8 \int {tanysec}^{2} ydy - 8 \int tanydy$ $8 [\frac{\tan^{2} y}{2}] - 8 \ln (secy)$ ${4 \tan}^{2} y - 4 \ln (\sec^{2} y) = x - 4 \ln (\frac{{4 \sec}^{2} y}{4})$ $x - 4 \ln (\frac{4 {t a n}^{2} y + 4}{4}) = x - 4 \ln (\frac{x + 4}{4})$ $x - 4 \ln ∣ x + 4 ∣ + 4 l n ∣ 4 ∣$ $x - 4 l n ∣ x + 4 ∣ + C$
	Answered by peter frank last updated on 24/Apr/24

	$u = x + 4 x = u - 4$ $du = dx$ $\int \frac{xdu}{u} = \int \frac{u - 4}{u} du = \int 1 du - \int \frac{4}{u} = u - 4 \ln u$ $\int 1 du - \int \frac{4}{u} du = u - 4 \ln u$ $x + 4 - 4 \ln (x + 4)$
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